Question

A solution is prepared by dissolving 27.75g Sulfuric Acid, H2So4 in enough water to make exactly 200. mL of solution. If Density = 1.1094g/mL What is the Molarity of H2S04? AND What is the Molality of H2S04?

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)= 27.75 g

use:

number of mol of H2SO4,

n = mass of H2SO4/molar mass of H2SO4

=(27.75 g)/(98.09 g/mol)

= 0.2829 mol

volume , V = 2*10^2 mL

= 0.2 L

use:

Molarity,

M = number of mol / volume in L

= 0.2829/0.2

= 1.415 M

Answer: 1.42 M

2)

mass of solution = density * volume

= 1.1094 g/mL * 200 mL

= 221.88 g

mass of solvent = mass of solution - mass of H2SO4

= 221.88 g - 27.75 g

= 194.13 g

= 0.19413 Kg

use:

Molality,

M = number of mol / mass of solvent in Kg

= 0.2829/0.19413

= 1.46 m

Answer: 1.46 m