Calculate molality of 1 litre solution of 93 % H2SO4 by volume. The density of solution is 1.84 g ml-1

Question

Calculate molality of 1 litre solution of 93 % H2SO4 by volume. The density of solution is 1.84 g ml-1 (A) 9.42 (B) 10.42 (C) 11.42 (D) 12.42

Tardigrade Admin · Accepted Answer

Given H2SO4 is 93% by volume wt. of H2SO4=93 g Volume of solution = 100 ml ∴ Weight of solution =100 × 1.84 g =184 g wt. of water = 184 - 93 = 91 g Molality =( textMoles/ textwt. of water in kg) =(93×1000/98×91)=10.42

Q. Calculate molality of $1$ litre solution of $93%$ $H_{2} S O_{4}$ by volume. The density of solution is $1.84 g m l^{- 1}$

$9.42$

$10.42$

$11.42$

$12.42$

Given $H_{2} S O_{4}$ is 93% by volume
wt. of $H_{2} S O_{4} = 93 g$
Volume of solution = 100 ml
$∴$ Weight of solution $= 100 \times 1.84 g = 184 g$
wt. of water $= 184 - 93 = 91 g$
Molality $= \frac{Moles}{wt. of water in kg}$
$= \frac{93 \times 1000}{98 \times 91} = 10.42$

Q. Calculate molality of 1 litre solution of 93% H2​SO4​ by volume. The density of solution is 1.84gml−1

9.42

10.42

11.42

12.42