Question

The molarity of 15% (w/w) solution of $H_2{SO}_4$ of density 1.1 g/cc is:

• A. $1.24M$
• B. $1.43M$
• C. $1.82M$
• D. $1.68M$

Answer 1

The correct option is D $1.68M$

We have $15\%$ weight by volume solution of $H_{2}S{O}_4$ of density $1.1$ g/cc.

This means we have $15$ g of $H_{2}S{O}_4$ in 100 g solution.

Moles of $H_{2}S{O}_4=\frac{15}{98}=0.153$ mol.
Density $=1.1$ g/mL

$⟹100$ g of total solution $=\frac{1}{1.1}\times 100=90.909$ mL.
$\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume}}$
$\text{Molarity}=\frac{15}{98}\times \frac{1000}{90.909}$ $=1.68M$

Hence, the molarity is $1.68M$.