Question

100g of 104.5 oleum is diluted with 94.5g h2o the molality of h2so4 in the resultant solution

Answer 1

This is a question testing the concept of molarity and molality

We first write down the equation :

H₂S₂O₇ + H₂O — > 2 H₂SO₄

The mole ratio is 1 : 2

100g of oleum is diluted.

Moles of oleum in 100g are :

Molar mass of oleum = 178

100 / 178 = 0.5618 moles.

Mole ratio is 1 : 2

The moles of H₂SO₄ is thus :

2 × 0.5618 moles = 1.1236 moles

Molality = moles of solute / Kg of solvent

Kg of solvent equals :

94.5 / 1000 = 0.0945 Kg

Molality = 1.1236 / 0.0945 = 11.89

11.89 m

Answer 2

Answer: The molality of $H_2SO_4$ in the solution is 11.89m.

Explanation: Moles is calculated by using the formula:

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of oleum will be:

Given mass = 100g

Molar mass of oleum = 178g/mol

Putting values in above equation, we get:

$\text{Moles of oleum}=\frac{100g}{178g/mol}=0.56moles$

Reaction of oleum to form sulfuric acid is given by:

$H_2S_2O_7+H_2O\rightarrow 2H_2SO_4$

By Stoichiometry,

1 mole of oleum produces 2 moles of sulfuric acid.

0.56 moles of oleum will produce = (2 × 0.56) = 1.12 moles of sulfuric acid.

Molality is calculated by using the formula:

$Molality=\frac{\text{Moles of solute}}{\text{Mass of solvent (in kg)}}$

Mass of solvent $(H_2O)$ = 94.5g = 0.0945kg     (Conversion factor: 1kg = 1000g)

Putting values in above equation, we get:

$Molality=\frac{1.12}{0.0945}=11.85m$