Question

What is the molarity of a $30$% (vol/vol) solution of $H_{2}S{O}_4$? (Molecular weight of $H_{2}S{O}_4=98$; density of $H_{2}S{O}_4=1.8g/ml$)

• A. $1.5M$
• B. $2.0M$
• C. $3.0M$
• D. $5.5M$

Answer 1

Answer: (D)

$5.5M$

$30\%\left(V/V\right)$ Solution of $H_{2}S{O}_4$ means

$30ml$ of $H_{2}S{O}_4$ is present in $100ml$ solution

So, in $V=100ml$, there is $(1.8\times 30)\text{grams}$ of $H_{2}S{O}_4$

$\Rightarrow$ For $V=100ml$, $54\text{grams}$ of $H_{2}S{O}_4$

Molarity $=\frac{M}{98}\times \frac{1000}{V}$$=\frac{54}{98}\times \frac{1000}{100}$$=\frac{540}{98}=5.5M$