Question

A solution is prepared by dissolving 50.7 g sucrose (C12H22O11) in 0.394 kg of water. The final volume of the solution is 355 mL. calculate:

a- molarity b- molality c- percent by mass d- mole fraction

Answer 1

a)

Molar mass of C12H22O11,

MM = 12*MM(C) + 22*MM(H) + 11*MM(O)

= 12*12.01 + 22*1.008 + 11*16.0

= 342.296 g/mol

mass(C12H22O11)= 50.7 g

use:

number of mol of C12H22O11,

n = mass of C12H22O11/molar mass of C12H22O11

=(50.7 g)/(3.423*10^2 g/mol)

= 0.1481 mol

volume , V = 3.55*10^2 mL

= 0.355 L

use:

Molarity,

M = number of mol / volume in L

= 0.1481/0.355

= 0.4172 M

Answer: 0.417 M

b)

m(solvent)= 0.394 Kg

use:

Molality,

m = number of mol / mass of solvent in Kg

=(0.1481 mol)/(0.394 Kg)

= 0.3759 molal

Answer: 0.376 molal

c)

% m/m = mass of sucrose * 100 / mass of solution

= 50.7 g *100 / (50.7 + 394) g

= 11.4 %

Answer: 11.4 %

d)

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

mass(H2O)= 0.394 Kg

= 394.0 g

use:

number of mol of H2O,

n = mass of H2O/molar mass of H2O

=(3.94*10^2 g)/(18.02 g/mol)

= 21.87 mol

Mole fraction = mol of sucrose / total mol

= 0.3759 / (0.3759 + 21.87)

= 0.0169

Answer: 0.0169