NCERT Solutions for Class 11 Science Chemistry Chapter 1 - Some Basic Concepts Of Chemistry

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Answer:

Precision refers to the closeness of various measurements for the same quantity. However, accuracy is the agreement of a particular value to the true value of the result. Results of A are very close to 3 g.

Hence, the correct answer is option (ii).

Answer:

The temperatures on the two scales are related to each other by the following relationship:


$°F=\frac{9}{5}(°C)+{32}_{}$

On substituting the values we have,

$$\begin{gathered} 200=\frac{9}{5}\left({ }^{°}C\right)+32 \\ 200-32=\frac{9}{5}\left({ }^{°}C\right) \\ 168=\frac{9}{5}\left({ }^{°}C\right) \\ \left({ }^{°}C\right)=168\times \frac{5}{9} \\ \left({ }^{°}C\right)=\frac{840}{9} \\ \left({ }^{°}C\right)=93.3 \end{gathered}$$


Hence, the correct answer is option (iii).

Answer:

 The formula for calculating molarity be given as:

$\mathrm{Molarity}=$$\frac{\mathrm{Moles} \mathrm{of} \mathrm{solute}}{\mathrm{Volume} \mathrm{of} \mathrm{solution}\left(\mathrm{L}\right)}$  

The given data is:

Mass of NaCl = $5.85g$
Volume of solution = $500 \mathrm{mL}$ = $500\times {10}^{-3}\mathrm{L}=0.5 \mathrm{L}$

$$\begin{gathered} \mathrm{No}. \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{NaCl}=\frac{\mathrm{given} \mathrm{mass}}{\mathrm{molar} \mathrm{mass}} \\ \mathrm{No}. \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{NaCl}=\frac{5.85}{58.5}=0.1 \mathrm{moles} \end{gathered}$$

Now, the molarity can be calculated as:

$\mathrm{Molarity}=\frac{0.1}{0.5}=0.2 \mathrm{mol} {\mathrm{L}}^{-1}$

Hence, the correct answer is option (iii).

Answer:

The molarity of the solution obtained can be calculated by using the following relation:

${\mathrm{M}}_1{\mathrm{V}}_1={\mathrm{M}}_2{\mathrm{V}}_2$

or  $$\begin{gathered} {\mathrm{M}}_2=\frac{{\mathrm{M}}_1{\mathrm{V}}_1}{{\mathrm{V}}_2} \\ {\mathrm{M}}_2=\frac{5\times 500}{1500}=1.66 \mathrm{M} \end{gathered}$$

Hence, the correct answer is option (ii). 

Answer:

 Let's calculate the moles for each species given.

We know that moles = $\frac{\mathrm{Given} \mathrm{mass}}{\mathrm{Molar} \mathrm{mass}}$

Applying this formula for calculating the respective moles of each species, we have:

Moles of 4g He = $\frac{4}{4}$ = 1 mole

Moles of 46g Na = $\frac{46}{23}$ = 2 moles

Moles of 0.40g Ca = $\frac{0.40}{40}$ = 0.01 moles

Moles of 12g He = $\frac{12}{4}$ = 3 moles

Since one mole of an element contains Avogadro number of atoms, therefore, the number of atoms will be maximum for 3 moles of He i.e.$3\times 6.022\times {10}^{23}$ atoms. 

Hence, the correct answer is option (iv).

Answer:

Molarity is defined as the number of moles of solute present in per litre of the solution. Thus, in order to find out the molarity of glucose, we have to calculate its moles.

In this question, 0.9 g of glucose is present in 1-litre solution. Therefore


$$\begin{gathered} \mathrm{moles}=\frac{\mathrm{given} \mathrm{mass}}{\mathrm{molar} \mathrm{mass}} \\ \mathrm{moles}=\frac{0.9 \mathrm{g}{ }^{}}{180 \mathrm{g} {\mathrm{mol}}^{-1}} \\ \mathrm{moles}=0.005 \mathrm{mole} \end{gathered}$$

Therefore, the molarity of glucose in the blood is 0.005 mol L^-1.

Hence, the correct answer is option (iii).

 

Answer:

Molality is defined as the number of moles of solute present in per kg of the solvent.

$\mathrm{molality}=\frac{\mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{mass} \mathrm{of} \mathrm{solvent} \mathrm{in} \mathrm{Kg}}$

$$\begin{gathered} \mathrm{moles} \mathrm{of} \mathrm{HCl}=\frac{18.25 \mathrm{g}}{36.5 \mathrm{g} {\mathrm{mol}}^{-1}}=0.5 \mathrm{mole} \\ \mathrm{mass} \mathrm{of} \mathrm{solvent}=500 \mathrm{g}=500\times {10}^{-3} \mathrm{Kg}=0.5 \mathrm{Kg} \\ \mathrm{Therefore}, \mathrm{molality}=\frac{0.5}{0.5}=1.0 \mathrm{m} \end{gathered}$$

Hence, the correct answer is option (iv).
 

Answer:

The moles can be calculated by using the formula for molarity as follows:

$$\begin{gathered} \mathrm{M}=\frac{\mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{Volume} \mathrm{of} \mathrm{solution} \mathrm{in} \mathrm{L}} \\ 0.02=\frac{\mathrm{moles} \mathrm{of} \mathrm{solute}}{100\times {10}^{-3}} \\ \mathrm{moles} \mathrm{of} \mathrm{solute}=0.002 \mathrm{mole} \\ 1 \mathrm{mole} \mathrm{of} \mathrm{a} \mathrm{substance} \mathrm{contain} 6.022\times {10}^{23} \mathrm{molecules}. \\ \mathrm{Therefore}, 0.002 \mathrm{moles} \mathrm{of} \mathrm{Acid} \mathrm{will} \mathrm{contain} 0.002\times 6.022\times {10}^{23} \mathrm{molecules}. \\ \mathrm{Thus} \mathrm{number} \mathrm{of} \mathrm{molecules} \mathrm{present} \mathrm{in} \mathrm{acid}=12.044\times {10}^{20}. \\ \mathrm{Hence}, \mathrm{the} \mathrm{correct} \mathrm{answer} \mathrm{is} \mathrm{option} \left(\mathrm{i}\right). \end{gathered}$$

Answer:

The mass percent of carbon in carbon dioxide can be calculated as follows:

$\mathrm{Mass} \mathrm{percent} \mathrm{of} \mathrm{C} \mathrm{in} {\mathrm{CO}}_2=\frac{\mathrm{Molar} \mathrm{mass} \mathrm{of} \mathrm{C}}{\mathrm{Molar} \mathrm{mass} \mathrm{of} {\mathrm{CO}}_2 }\times 100$


$T\mathrm{herefore}, ⁒ \mathrm{of} \mathrm{C} \mathrm{in} {\mathrm{CO}}_2=\frac{12}{44}\times 100=27.27⁒$

Hence, the correct answer is option (ii).

Answer:

Empirical formula = CH₂O

Empirical formula mass = 12+2+16 = 30 g

The molecular mass of the compound = 180 g

We know that molecular formula = n x empirical formula

 where n is the common factor or multiplying factor and can have the values 1,2,3......etc.

Therefore, n = $\frac{\mathrm{molecular} \mathrm{formula} mass}{\mathrm{empirical} \mathrm{formula} mass}=\frac{180}{30}=6$

Thus, the molecular formula of the compound = (CH₂O)n = (CH₂O)6 = C₆H₁₂O₆

Hence, the correct answer is option (iii).

 

Answer:

$\mathrm{Density}=\frac{\mathrm{Mass}}{\mathrm{Volume}}$

The given density and mass are  3.12 g mL^–1 and 1.5 mL respectively. Now, substituting these values in the above formula, we have:

$$\begin{gathered} 3.12 \mathrm{g} {\mathrm{mL}}^{–1}=\frac{\mathrm{Mass}}{1.5 \mathrm{mL}} \\ \mathrm{Mass}= 3.12 \mathrm{g} {\mathrm{mL}}^{–1}\times 1.5 \mathrm{mL} \\ \mathrm{Mass}=4.680 \mathrm{g} \end{gathered}$$

Since the final result should be reported up to the same number of significant figures as present in the least precise number ( here two significant figures are there in 1.5), Therefore the final answer is to be calculated up to 2 significant figures. It means in 4.680, the digit 8 is to be removed.
Since 6 is greater than 5, thus 1 is to be added to the previous digit i.e. 6.

Therefore, the correct value is 4.7

Hence, the correct answer is option (i).

 

Answer:

The compound doesn't retain the physical properties of its constituent elements.

Hence, the correct answer is option (iii).

Answer:

According to the law of mass conservation, the mass of reactants in a chemical reaction is equal to the mass of products.

Hence, the correct answer is option (i).

Answer:

The number of atoms on the reactant side should be equal to the number of atoms on the product side or the reaction should be balanced. 

This is not satisfied by (ii) and thus is incorrect.

Hence, the correct answer is option (ii).

Answer:

According to the law of multiple proportion, if two elements combine to form more than one compound, the masses of one element that combine with the fixed mass of another element, are in the ratio of small whole numbers.

Hence, the correct answer is option (ii).

Answer:

We know that 1 mole of oxygen at STP = 6.022 × 10²³ molecules of oxygen

Molar mass of oxygen gas = 32 g mol^-1
∴ mass of 1 mole of oxygen gas = 32 g

Hence, the correct answers are options (i) and (iv).
 

Answer:

0.1M sulphuric acid solution means 0.1 mole of sulphuric acid in one litre of the solution.

Therefore moles of ${\mathrm{H}}_2{\mathrm{SO}}_4$ = 0.1 

Similarily moles of sodium hydroxide in 0.1 M solution = 0.1

Now, according to the reaction, 2 moles of NaOH react with 1 mole ${\mathrm{H}}_2{\mathrm{SO}}_4$.
 Therefore 1 mole of NaOH reacts with ($\frac{1}{2}=0.5$ moles) of ${\mathrm{H}}_2{\mathrm{SO}}_4$.

Therefore, 0.1 mole of NaOH reacts with $0.5\times 0.1=0.05$ moles. 

It means that all of the NaOH will react and thus, act as a limiting reagent.

Now, according to the reaction:

2 moles of NaOH reacts to form 1 mole of Na₂SO₄

Therefore, 0.1 mole NaOH = $\frac{1}{2}\times 0.1=0.05$  moles of Na₂SO₄
Thus, mass of Na₂SO₄ formed = moles × molar mass = $0.05 \mathrm{mol}\times 142.04 \mathrm{g} {\mathrm{mol}}^{-1}=7.10 \mathrm{g}$ 

The molarity of Na₂SO₄ may be calculated as follows:

Molarity = $\frac{\mathrm{moles}}{\mathrm{volume} \mathrm{of} \mathrm{solution} \mathrm{in} \mathrm{L}}=\frac{0.05}{2}=0.025 \mathrm{M}$  

( since the total volume will become 2 L )

Hence, the correct answers are options (ii) and (iii). 

Answer:

Since one mole of an atom contains $6.022\times {10}^{23}$ atoms. Therefore we have to calculate the number of atoms for the respective case.

We know that moles $=\frac{\mathrm{given} \mathrm{mass}}{\mathrm{molar} \mathrm{mass}}$

(i)  16 g oxygen = $\frac{16}{32}=$0.5 moles of oxygen = $0.5\times 6.022\times {10}^{23}$ atoms = $3.011\times {10}^{23}$ atoms

      4 g  H₂ = $\frac{4}{2}=$2 moles of hydrogen = $2\times 6.022\times {10}^{23}$ atoms = $12.044\times {10}^{23}$ atoms

(ii) 16 g oxygen = $3.011\times {10}^{23}$ atoms

      44 g CO₂ = $\frac{44}{44}=$1.0 mole of CO₂ =  $6.022\times {10}^{23}$ atoms

(iii) 28 g N₂ = $\frac{28}{28}=$1.0 mole of N₂ =  $6.022\times {10}^{23}$ atoms

       32 g  O₂ = $\frac{32}{32}=$1.0 mole of O₂ =  $6.022\times {10}^{23}$ atoms

(iv) 12 g C = $\frac{12}{12}=$1.0 mole of C =  $6.022\times {10}^{23}$ atoms

       23 g Na = $\frac{23}{23}=$1.0 mole of Na = $6.022\times {10}^{23}$ atoms

It is clear from the calculations that the number of atoms is the same in (iii) and (iv).


Hence, the correct answers are options (iii) and (iv). 


 

Answer:

We know that the molarity can be calculated as follows:

$\mathrm{Molarity} =$ $\frac{\mathrm{gram} \mathrm{mass} \mathrm{of} \mathrm{solute}\times 1000}{\mathrm{Molar} \mathrm{mass} \mathrm{of} \mathrm{solute}\left({\mathrm{M}}_{\mathrm{B}}\right)\times \mathrm{Volume} \mathrm{of} \mathrm{solution} \mathrm{in} \mathrm{mL}}$    OR   $\mathrm{Molarity}=\frac{\mathrm{moles} \mathrm{of} \mathrm{solute}\times 1000}{\mathrm{volume} \mathrm{of} \mathrm{solution} \mathrm{in} \mathrm{mL}}$

Now, let's calculate the concentration for each case:

(i)  $\mathrm{Molarity} =$ $\frac{20\times 1000}{40\times 200}=2.5 \mathrm{M}$ ;${\mathrm{M}}_{\mathrm{B}} = 40 \mathrm{g} {\mathrm{mol}}^{-1}$

(ii) $\mathrm{Molarity} =$ $\frac{0.5\times 1000}{200}=2.5 \mathrm{M}$ 

(iii) $\mathrm{Molarity} =$ $\frac{40\times 1000}{40\times 100}=10 M$; ${\mathrm{M}}_{\mathrm{B}} = 40 \mathrm{g} {\mathrm{mol}}^{-1}$


(iv) $\mathrm{Molarity} =$ $\frac{20\times 1000}{56\times 200}=1.78 \mathrm{M}$; ${\mathrm{M}}_{\mathrm{B}} = 56 \mathrm{g} {\mathrm{mol}}^{-1}$

The concentrations are the same in (i) and (ii).

Hence the correct answers are options (i) and (ii).

 

Answer:

We know that,

1 mole = $6.022\times {10}^{23}$ molecules ; $\mathrm{Moles}=\frac{\mathrm{given} \mathrm{mass}}{\mathrm{molar} \mathrm{mass}}$


Therefore, the number of molecules in 16 g oxygen = $\mathrm{No}. \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{oxygen}\times 6.022\times {10}^{23}$ molecules
                                                                                   = $\frac{16}{32}\times 6.022\times {10}^{23}=3.011\times {10}^{23}$molecules


(i)   the number of molecules in 16 g CO = $\frac{16}{28}\times 6.022\times {10}^{23}=0.0948\times \times {10}^{23}$ molecules

(ii)  the number of molecules in 28 g N₂ = $6.022\times {10}^{23}$ molecules

(iii) the number of molecules in 14 g N₂ = $\frac{14}{28}\times 6.022\times {10}^{23}=3.011\times {10}^{23}$ molecules

(iv) the number of molecules in 1.0 g H₂ = $\frac{1}{2}\times 6.022\times {10}^{23}=3.011\times {10}^{23}$ molecules

Thus, (iii) and (iv) have the same number of molecules as oxygen.

Hence, the correct answers are options (iii) and (iv).




 

Answer:

 We know that,

Mass percent = $\frac{\mathrm{Mass} \mathrm{of} \mathrm{solute}}{\mathrm{Mass} \mathrm{of} \mathrm{solution}}\times 100$

Mole fraction: It is the ratio of the number of moles of a particular component to the total number of moles of the solution.

Since both mass percent and mole fraction are unitless, Hence, the correct answer is option (iii) and (iv).

 

Answer:

Law of conservation of mass states that matter can neither be created nor be destroyed.

Avogadro proposed that an equal volume of the gases at the same temperature and pressure should contain an equal number of molecules.

Hence, the correct answers are options (i) and (iv).

Answer:

We know that,

Mass of $6.022\times {10}^{23}$ atoms of carbon in gram = 12 g

Therefore, the mass of one atom of carbon in gram = $\frac{12}{6.022\times {10}^{23}}=1.993\times {10}^{-23}$g

Answer:

We know that in multiplication or division, the final result should be reported upto the same number of significant figures as present in the leas precise number ( Here 2 significant figures ).


$\frac{2.5\times 1.25\times 3.5}{2.01}=\frac{10.9375}{2.01}=5.44154$

Therefore, two significant figures should be present in the above calculation.
 

Answer:

The symbol for the SI unit of mole is mol.

One mole is defined as the amount of substance that contains as many particles or entities as there are atoms in exactly 12 g of the ¹²C isotope.

 

Answer:

Molarity is the number of moles of solute present in per litre of the solution i.e.

Molarity = $\frac{\mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{volume} \mathrm{of} \mathrm{solution} \left(\mathrm{L}\right)}$


Molality is the number of moles of solute present in per kg of the solvent i.e.

Molality = $\frac{\mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{mass} \mathrm{of} \mathrm{solvent} \left(\mathrm{kg}\right)}$


Molality is independent of temperature whereas molarity is temperature dependent.

Answer:

The mass percent of the respective species in Ca₃(PO₄)₂ may be calculated as follows:


Mass % of Calcium = $\frac{3\times (\mathrm{atomic} \mathrm{mass} \mathrm{of} \mathrm{calcium})}{\mathrm{molecular} \mathrm{mass} \mathrm{of} {\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2}\times 100=\frac{3\times 40}{310}\times 100=38.71\%$

Mass % of phosphorous = $\frac{2\times (\mathrm{atomic} \mathrm{mass} \mathrm{of} \mathrm{P})}{\mathrm{Molecular} \mathrm{mass} \mathrm{of} {\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2}\times 100=\frac{2\times 31}{310}\times 100 = 20 \%$

Mass % of oxygen = $\frac{8\times (\mathrm{mass} \mathrm{of} \mathrm{oxygen})}{\mathrm{molecular} \mathrm{mass} \mathrm{of} {\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2}\times 100= \frac{2\times 16}{310}\times 100= 41.29 \%$

Answer:

Gay Lussac's law of gaseous volumes is being obeyed in this experiment. According to this law when gases combine or are produced in a chemical reaction, they do so in a simple ratio by volume, provided all the gases are at the same temperature and pressure. 

In this case, all the gases are in a simple ratio (2:1:2).

Answer:

(a) Yes, the given statement is true.

(b) This is in accordance with the law of multiple proportions.

(c) Carbon combines with oxygen to form two compounds, CO₂ and CO.
 

Answer:

The average atomic mass of hydrogen can be calculated as follows:


Average atomic mass = $\frac{(\mathrm{Natural} \mathrm{abundance} \mathrm{of} {}^1\mathrm{H}\times \mathrm{molar} \mathrm{mass} \mathrm{of} {}^1\mathrm{H})+(\mathrm{Natural} \mathrm{abundance} \mathrm{of} {}^2\mathrm{H}\times \mathrm{molar} \mathrm{mass} \mathrm{of} {}^2\mathrm{H})}{100}$

                                   = $$\begin{gathered} \frac{(99.985\times 1)+(0.015\times 2)}{100}=\frac{ \\ 100.015}{100}=1.00015 \mathrm{u} \end{gathered}$$

Answer:

According to the given data:

63.5 g of Zn reacts with HCl to form 22.7 L of H₂

Therefore, 1.0 g of Zn reacts with HCl to form $\frac{22.7}{63.5}$ L of H₂

Thus, 32.65 g of Zn reacts with HCl to form $\frac{22.7}{65.3}\times 32.65=11.67 \mathrm{L}$ of H₂

 

Answer:

3 molal solution of NaOH means 3 mol of NaOH is dissolved in 1000 g of the solvent.

Therefore, the mass of NaOH = $\mathrm{moles}\times \mathrm{molar} \mathrm{mass}=3\times 40=120 \mathrm{g}$

We know that,

mass of solution = mass of solute + mass of solvent
                           = 120 + 1000 
                           = 1120 g

$\mathrm{volume} \mathrm{of} \mathrm{solution}=\frac{\mathrm{mass} \mathrm{of} \mathrm{solution}}{\mathrm{density} \mathrm{of} \mathrm{solution}} ( \mathrm{since}, \mathrm{density}=\frac{\mathrm{mass}}{\mathrm{volume}})$

                           $=\frac{1120}{1.110} \mathrm{mL}=1009 \mathrm{mL}$

Now, Molarity = $\frac{\mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{volume} \mathrm{of} \mathrm{solution}}=\frac{3}{1.009}=2.973 \mathrm{M}$

Answer:

We know that,

Molality = $\frac{\mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{mass} \mathrm{of} \mathrm{solvent} \left(\mathrm{kg}\right)}$

Since mass doesn't depend upon temperature, therefore, molality is independent of temperature.

Answer:

According to the given data:

mass of NaOH = 4 g

moles of NaOH (n) = $\frac{4 \mathrm{g}}{40 \mathrm{g} {\mathrm{mol}}^{-1}}=0.1 \mathrm{mol}$

mass of H₂O = 36 g

moles of H₂O (n) = $\frac{36 \mathrm{g}}{18 \mathrm{g} {\mathrm{mol}}^{-1}}=2 \mathrm{mol}$

Now, mole fraction of NaOH = $\frac{{\mathrm{n}}_{\mathrm{NaOH}}}{{\mathrm{n}}_{\mathrm{NaOH}}+{\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}}=\frac{0.1}{0.1+2}=\frac{0.1}{2.1}=0.0476$

         mole fraction of H₂O = $\frac{{\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}}{{\mathrm{n}}_{\mathrm{NaOH}}+{\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}}=\frac{2}{0.1+2}=\frac{2}{2.1}=0.952$

Mass of solution = mass of water + mass of NaOH = 36 g + 4 g = 40 g

Volume of solution = $\frac{\mathrm{mass} \mathrm{of} \mathrm{solution}}{\mathrm{density} \mathrm{of} \mathrm{solution}}=\frac{40 \mathrm{g}}{1 \mathrm{g} {\mathrm{mL}}^{-1}}=40 \mathrm{mL}$


Therefore, molarity of solution = $\frac{\mathrm{no}. \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{volume} \mathrm{of} \mathrm{solution} \left(\mathrm{L}\right)}=\frac{0.1}{40\times {10}^{-3}}=2.5 \mathrm{M}$

Answer:

According to the given reaction, 2 moles of 'A' requires 4 moles of 'B' to react.

(i)  Therefore, for 5 moles of 'A', the moles of 'B' required = $\frac{4 \mathrm{moles} \mathrm{of} \text{'}\mathrm{B}\text{'}}{2 \mathrm{moles} \mathrm{of} \text{'}\mathrm{A}\text{'}}\times 5 \mathrm{moles} \mathrm{of} \text{'}\mathrm{A}\text{'}=10 \mathrm{moles} \mathrm{of} \text{'}\mathrm{B}\text{'}$

But, we have only 6 moles of 'B', therefore, 'B' is the limiting reagent.

(ii)  The amount of 'C' formed is determined by amount of 'B'.

Since 4 moles of 'B' gives 3 moles of 'C'. Hence 6 moles of 'B' will give $\frac{3 \mathrm{moles} \mathrm{of} \text{'}\mathrm{C}\text{'}}{4 \mathrm{moles} \mathrm{of} \text{'}\mathrm{B}\text{'}}\times 6 \mathrm{moles} \mathrm{of} \text{'}\mathrm{B}\text{'}=4.5 \mathrm{moles} \mathrm{of} \text{'}\mathrm{C}\text{'}.$
 

Answer:

(i) Moles of CO₂ = $\frac{88 g}{44 g mo{l}^{-1}}=2 mol$

(ii) The number of moles of H₂O in $6.022\times {10}^{23}$ molecules may be calculated as follows:

      Number of Moles = $\frac{\mathrm{No}. \mathrm{of} \mathrm{molecules}}{\mathrm{Avagadro} \mathrm{Number}0}=\frac{6.022\times {10}^{23}}{6.022\times {10}^{23}}=1 \mathrm{mol}$

(iii)  The number of moles at STP can be calculated as follows:

      Number of Moles = $\frac{\mathrm{given} \mathrm{volume}}{\mathrm{volume} \mathrm{at} \mathrm{STP}}=\frac{5.6}{22.4}=0.25 \mathrm{mol}$

(iv) The number of moles of O₂ = $\frac{96 \mathrm{g}}{32 \mathrm{g} {\mathrm{mol}}^{-1}}=3 \mathrm{mol}$

(v) We know that 1 mol of any gas contains $6.022\times {10}^{23}$ molecules.


Therefore, the correct matching is:   (i)$\to$(b) ; (ii)$\to$(c) ; (iii)$\to$(a) ; (iv)$\to$(e) ; (v)$\to$(d)
       

Answer:

The correct matching may be done as follows:

(i)$\to$(e) ; (ii)$\to$(d) ; (iii)$\to$(b) ; (iv)$\to$(g) ; (v)$\to$(c) ; (vi)$\to$(f) ; (vii)$\to$(a) ; (viii)$\to$(i)

Answer:

We know that the empirical formula of ethene is CH₂.

Thus, the empirical formula mass = 12 u + 2×1 u = 14 u.

The molecular mass of ethene = 28 u.

Therefore, it's clear that the empirical formula mass is half of the molecular mass.
Also, the empirical formula  shows that carbon and hydrogen are in the ratio 1:2 i.e.

(C : H) = 1:2

Therefore, both assertion and reason are correct and the reason is the correct explanation of assertion.

Hence the correct answer is option (i).



 

Answer:

Carbon-12 is one of the isotopes of carbon and can be represented as ¹²C. In this system,  ¹²C is assigned a mass of exactly 12 atomic mass units (amu) and mass of all other atoms are given relative to this standard.

One atomic mass unit is defined as a mass exactly equal to one-twelfth the mass of one carbon-12 atom.

Both the above statements are true but reason for the assertion is wrong.

Hence, the correct answer is option (ii).

 

Answer:

According to the rules for significant figures:

Zero at the end or right of a number are significant provided they are on the right side of the decimal point.

Therefore, the correct answer is option (iii).

 

Answer:

The combustion reaction of methane may be given as follows: 

CH₄(g) + 2 O₂(g) CO₂(g) + 2 H₂O(g)

From the reaction, it's clear that 1 mol of methane gives 2 mol of water i.e. 16 g of methane gives 36 g of water.

∵ moles of water = 2

∴ mass of water = 

As we can see that water is one of the products of the combustion reaction.

Hence, the correct answer is option (iii).

 

Answer:

According to the given data,

(i) 32 g of oxygen occupies 22.4 L of the volume at STP

∴ 1.6 g of oxygen will occupy,  $\frac{22.4 \mathrm{L}}{32 \mathrm{g}}\times 1.6 \mathrm{g}= 1.12 \mathrm{L}$ of the volume.

∴  P₁ = 1 atm ;  T₁ = 273.15K ;  V₁ = 1.12 L

Now, according to the given information, 

  P₂ = $\frac{ {\mathrm{P}}_1}{2}=\frac{1}{2}=0.5 \mathrm{atm}$ ;   V₂ = ?

Using Boyle's law,  P₁V₁ = P₂V₂

∴ ${\mathrm{V}}_2=\frac{{\mathrm{P}}_1{\mathrm{V}}_1}{{\mathrm{P}}_2}=\frac{1\mathrm{atm}\times 1.12 \mathrm{L}}{0.5 \mathrm{atm}}=2.24 \mathrm{L}$


(ii)   32 g of the oxygen contains molecules = $6.022\times {10}^{23}$ 

        1.6 g of the oxygen contains molecules = $\frac{6.022\times {10}^{23}}{32}\times 1.6= 3.011\times {10}^{22}$

Answer:

We know that,

$\mathrm{Molarity}=\frac{\mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{Volume} \mathrm{of} \mathrm{solution} \left(\mathrm{L}\right)}$

Thus, the moles of HCl can be calculated as follows:

Moles = $\mathrm{Molarity}\times \mathrm{Volume}=0.76 \mathrm{M}\times 250\times {10}^{-3} \mathrm{L} =0.19 \mathrm{mol}$

Now, moles of CaCO₃ = $\frac{1000 \mathrm{g}}{100 \mathrm{g} {\mathrm{mol}}^{-1}}=10 \mathrm{mol}$

According to the reaction given,

∵ 1 mol of CaCO₃ requires 2 mol of HCl
∴ 10 mol of CaCO₃ requires moles of HCl= $2\times 10=20 \mathrm{mol}$

But the available moles of HCl = 0.19 mol

Therefore, HCl is behaving as a limiting reagent. Thus, the amount of CaCl₂ formed will be calculated by the moles of HCl available.

∵ 2 mol of HCl gives CaCl₂ = 1 mol
∴ 0.19 mol of HCl gives CaCl₂ = $\frac{1}{2}\times 0.19=0.095 \mathrm{mol}$

∴ mass of CaCl₂ = $\mathrm{moles}\times \mathrm{molar} \mathrm{mass}=0.095 \mathrm{mol}\times 111 \mathrm{g} {\mathrm{mol}}^{-1}=10.545 \mathrm{g}$

 

Answer:

Answer:

According to the law of multiple proportions, when two elements combine to form two or more compounds, then the different masses of one element, which combine with a fixed mass of the other, bear a simple ratio to one another.

For the combination, AB

1g of A combines with $\frac{5}{2}$ g of B = 2.5 g B

For AB₂

1g of A combines with $\frac{10}{2}$ g of B = 5 g of B

For A₂B

1g of A combines with $\frac{5}{4}$ g of B = 1.25 g of B

For A₂B₃

1g of A combines with $\frac{15}{4}$ g of B = 3.75 g of B

Thus, it is proved that the law of multiple proportions is applicable.