Question

The molarity of 98% ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(d=1.8g/mL\right)$ by weight is:

• A. 6M
• B. 18M
• C. 10M
• D. 4M

Answer 1

The correct option is B 18M

Let us assume, we have 100 g solution

$98$ $gm$ $H_{2}S{O}_4$ is present in $100$ $g$ solution.

$Volumeofsolution=\frac{Mass}{Density}=\frac{100}{1.8}$ $mL$

Moles of $H_{2}S{O}_4=\frac{98}{98}=1$ $mole$

$Molarity=\frac{Moleofsolute}{Volume\left(Litres\right)}=\frac{1\times 1000}{100/1.8}=18$ $M$

Hence, the correct option is $\left(B\right)$.