Question

${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\left(\text{98g}{\text{mol}}^{\text{- 1}}\right)$ is 98% by mass of solution. If the density is 1.8 g/ml, the molarity is:

• A. 18M
• B. 36M
• C. 54M
• D. 0.18M

Answer 1

The correct option is A 18M

Given, $98$ $gm$ $H_{3}P{O}_4$ in $100$ $gm$ solution

Volume of solution $=\frac{100}{1.8}=\frac{1000}{18}$ $mL=\frac{1}{18}$ Litres

Moles of $H_{3}P{O}_4=\frac{98}{98}=1$

$Molarity=\frac{Molesofsolute}{V\left(solutioninlitres\right)}=\frac{1}{\frac{1}{18}}=18$ $M$