Question

A given solution of $H_{2}S{O}_4$ is labelled as 49% (w/w), then the correct statement regarding the solution is $(d=1.3\text{g/mL})$

• A. $m=\frac{500}{51}$
• B. $N=\frac{1000}{51}$
• C. $\%\text{w/v}=(49\times 1.3)\%$
• D. $M=6.5$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $m=\frac{500}{51}$
C $\%\text{w/v}=(49\times 1.3)\%$
D $M=6.5$

Option A
Let the mass of the solution be 100g
molality $=\frac{49}{98}\times \frac{1000}{51}=\frac{500}{51}$
Option B
For 100 ml solution,
Normality $=\frac{equivalents}{volume}=\frac{49}{49}\times \frac{1000}{100}\times 1.3$
$=13$
Option C
For 100 ml solution, mass of $H_{2}S{O}_4$ present:
$=49\times 1.3$
$\%\text{w/v}=(49\times 1.3)\%$
Option D
Molarity $=\frac{49\times 1.3}{98}\times \frac{1000}{100}=6.5$