Question

Calculate the molality of each of the following solutions.

1) 583g of H2SO4 in 1.50 kg of water.

2) 0.86g of NaCl in 1.00x10^2g of water.

3) 46.85g of codeine, C18H21NO3, in 125.5g of ethanol, C2H5OH.

4) 25g of I2 in 125g of ethanol, C2H5OH.

Answer 1

1)

Molar mass of H2SO4,

MM = 2*MM(H) + 1*MM(S) + 4*MM(O)

= 2*1.008 + 1*32.07 + 4*16.0

= 98.086 g/mol

mass(H2SO4)= 583 g

use:

number of mol of H2SO4,

n = mass of H2SO4/molar mass of H2SO4

=(5.83*10^2 g)/(98.09 g/mol)

= 5.944 mol

m(solvent)= 1.50 Kg

use:

Molality,

m = number of mol / mass of solvent in Kg

=(5.944 mol)/(1.5 Kg)

= 3.963 molal

Answer: 3.96 molal

2)

Molar mass of NaCl,

MM = 1*MM(Na) + 1*MM(Cl)

= 1*22.99 + 1*35.45

= 58.44 g/mol

mass(NaCl)= 0.86 g

use:

number of mol of NaCl,

n = mass of NaCl/molar mass of NaCl

=(0.86 g)/(58.44 g/mol)

= 1.472*10^-2 mol

m(solvent)= 100 g

= 0.1 kg

use:

Molality,

m = number of mol / mass of solvent in Kg

=(1.472*10^-2 mol)/(0.1 Kg)

= 0.1472 molal

Answer: 0.15 molal

3)

Molar mass of C18H21NO3,

MM = 18*MM(C) + 21*MM(H) + 1*MM(N) + 3*MM(O)

= 18*12.01 + 21*1.008 + 1*14.01 + 3*16.0

= 299.358 g/mol

mass(C18H21NO3)= 46.85 g

use:

number of mol of C18H21NO3,

n = mass of C18H21NO3/molar mass of C18H21NO3

=(46.85 g)/(2.994*10^2 g/mol)

= 0.1565 mol

m(solvent)= 125.5 g

= 0.1255 kg

use:

Molality,

m = number of mol / mass of solvent in Kg

=(0.1565 mol)/(0.1255 Kg)

= 1.247 molal

Answer: 1.247 molal

4)

Molar mass of I2 = 253.8 g/mol

mass(I2)= 25 g

use:

number of mol of I2,

n = mass of I2/molar mass of I2

=(25 g)/(2.538*10^2 g/mol)

= 9.85*10^-2 mol

m(solvent)= 125 g

= 0.125 kg

use:

Molality,

m = number of mol / mass of solvent in Kg

=(9.85*10^-2 mol)/(0.125 Kg)

= 0.788 molal

Answer: 0.79 molal