Ans. 1. Given, strength of sulfuric acid = 3.75 M
Density = 1.230 g/mL
Now,
3.75 M mean 3.75 mol of sulfuric acid in 1.0 L os solution.
Mass of H2SO4 in solution = Moles x molar mass
= 3.75 mol x (98.07948 g/mol)
= 367.79805 g
Mass of 1.0 L solution = Volume x density
= 1.0 L x (1.230 g/mL)
= 1230.0 g
= 1.230 kg
#1. A. Mass % of H2SO4 = (mass of H2SO4/ Mass of solution) x 100
= (367.79805 g / 1230.0 g)
= 29.90 %
#1. B. Molality of H2SO4 = moles of H2SO4/ mass of solvent (i.e. water) in kg
= 3.75 mol / 1.230 kg
= 3.05 mol/kg
= 3.05 m
Ans. 2. Ans. Freezing point depression, dTf is given by-
dTf = i Kf m - equation 1
where, i = Van’t Hoff factor = 1 for non-dissociating solutes
Kf = molal freezing point depression constant of the solvent
m = molality of the solution
dTf = Freezing point of pure solvent – Freezing point of solution
= depression in freezing point.
Let the molar mass of the compound be Y g/mol.
So, number of moles of the compound = mass/ molar mass
= 0.500 g/ (Y g/mol)
= (0.50 / Y) mol
Molality of solution = Moles of solute / Mass of solvent in kg
= (0.50 / Y) mol / 0.025 kg ; [1 kg = 1000 g]
= (20/ Y) mol/ kg
= (20/ Y) m
Given,
Depression in freezing point, dTf = 8.680C
Kf = 40.00C/ m
Putting the values in equation 1-
8.680C = 1 x (40.00C/ m) x (20/ Y) m
Or, 8.680C = 1 x (40.00C/ m) x (20/ Y) m
Or, 8.680C = 8000C/ Y
Or, Y = 8000C / 8.680C = 92.165
Hence, molar mass of the compound = Y g/mol = 92.165 g/mol
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