Question

Calculate the molality and molarity of a solution made by mixing equal volumes of

30 %

by weight of

H_{2} {SO}_{4}

(density

= 1.20 g / mL

) and

70 %

by weight of

H_{2} {SO}_{4}

(density

= 1.60 g / mL

).

Accepted Answer

Given,

Two Solutions:

Sol A = $30 %$ (w/W) of density = $1.20 g / mL$ of $H_{2} {SO}_{4}$ .

Sol B = $70 %$ (w/W) of density = $1.60 g / mL$ of $H_{2} {SO}_{4}$ .

Molality of solution is the number of moles of solute mass of solvent $(Kg)$

Molarity of solution is the number of moles of solute volume of solution $(L)$

Mass of solvent = Mass of solution − Mass of solute.

For A $= (volume \times density)$ of solution $- 30 %$ by weight of solution $= 100 \times 1.20 - 0.3 \times 120 = 84 g$ .

For B $= (volume \times density)$ of solution $- 70 %$ by weight of solution $= 100 \times 1.60 - 0.7 \times 160 = 48 g$ .

$Molality of solution A + B = \frac{no . of moles of solute A}{mass of solvent (Kg)} + \frac{no . of moles of solute B}{mass of solvent (Kg)}$

$= \frac{36 / 98 + 112 / 98}{84 + 48} \times 1000 = 11.44 m .$

Let, the volume of each solution is $100 mL$ .

$Molarity of A + B = \frac{no . of moles of solute A}{volume of solution (L)} + \frac{no . of moles of solute B}{volume of solution (L)}$

$= \frac{39 / 98 + 112 / 98}{200} \times 1000 = 7.55 M .$

Molality $= 11.44 m$ , Molarity $= 7.55 M$

Solute	Dissociation/association reaction	Degree of dissociation/association	n	i
$KCl$		$1$
$H_{2} {SO}_{4}$		$1$
${CH}_{3} COOH$ (in water)		$0.2$
${CH}_{3} COOH$ (in benzene)		$0.5$
Urea
$NaBr$		$0.8$
A	$3 A \to A_{3}$	$1$

Embibe Experts Solutions for Chapter: Solutions, Exercise 1: Exercise-1

Embibe Experts Chemistry Solutions for Exercise - Embibe Experts Solutions for Chapter: Solutions, Exercise 1: Exercise-1

Questions from Embibe Experts Solutions for Chapter: Solutions, Exercise 1: Exercise-1 with Hints & Solutions

Learn and Prepare for any exam you want !