a 0.65M BaCl2 solution is prepared by dissolving pure solid bacl2. 2h2o in water. determine the mass of hydrated salt dissolved per milliliter of solution and mass of anhydrous bacl2 present per milliliter of solution.
Dear student
One millilitre of the solution = 10-3 litre of solution.
⇒ Moles of hydrated salt required for 1.0 mL solution =0.65 × 10^3
⇒ Mass of hydrated salt required for 1.0 mL solution = Moles × Molar mass
= 0.65×10−3×244=0.1586 g
Also, 1.00 moles of hydrated BaCl 2.2H2O gives 1.00 moles of anhydrous BaCl2 in solution:
Moles of anhydrous BaCl2 per mL of solution = 0.65×10^-3
⇒ Mass of anhydrous BaCl2 present in 1.0 mL solution = 0.65×10^-3×208
= 0.1352 g (208 is the molar mass of anhydrous BaCl2 ).
Regards
One millilitre of the solution = 10-3 litre of solution.
⇒ Moles of hydrated salt required for 1.0 mL solution =0.65 × 10^3
⇒ Mass of hydrated salt required for 1.0 mL solution = Moles × Molar mass
= 0.65×10−3×244=0.1586 g
Also, 1.00 moles of hydrated BaCl 2.2H2O gives 1.00 moles of anhydrous BaCl2 in solution:
Moles of anhydrous BaCl2 per mL of solution = 0.65×10^-3
⇒ Mass of anhydrous BaCl2 present in 1.0 mL solution = 0.65×10^-3×208
= 0.1352 g (208 is the molar mass of anhydrous BaCl2 ).
Regards