Question

An aqueous solution contains 167g of CuSo4 in 820cm3 of solution. The density of the
solution is 1.95g/cm3 . calculate:-
(1) Formality
(2) Molarity
(3) Percentage by mass
(4)Mole fraction
(6) Molality of solution​

Answer 1

Answer:

Explanation:

What is the molarity?

Mass of CuSO4 in 1,000.0 mL solution = 1000.0 mL / 820.0 mL * 167.0 g = 203.66 g

Molar mass CuSO4 = 159.6 g/mol

Mol CuSO4 in 203.66 g = 203.66 g / 159.6 g/mol = 1.28 mol in 1.00 L solution

Molarity = 1.28 M

percent by mass?

Mass of 820 mL solution = 820 mL * 1.195 g/mL = 979.9 g

% by mass = 167.0 g / 979.9 g * 100 % = 17.04 m/m

molar fraction?

The solution consists of 167.0 g CuSO4 and 979.9 g - 167.0 g = 812.9 g H2O

Mol CuSO4 in 167.0 g = 167.0 g / 159.6 g/mol = 1.046 mol

Mol H2O in 812.9 g H2O = 812.9 g/18.0 g/mol 45.161 mol H2O

Total moles = 1.046 + 45.161 = 46.207 mol

Mol fraction CuSO4 = 1.046 / 46.207 = 0.0225

Mol fraction H2O = 0.9775

Note : mol fraction has no units . Sum of mol fractions = 1.0

and molality?

We previouly calcuated that the solution contains:

167.0 g CuSO4 and 979.9 g - 167.0 g = 812.9 g H2O

Mass CuSO4 in 1000 g water = 1000 g / 812.9 g * 167g CuSO4 = 205.66 g CuSO4 in 1.0 kg water

Mol CuSO4 in 205.66 g = 205.66 / 159.6 g/mol = 1.29 mol /kg water

Molality = 1.29 m