Question

The molarity of $98\%$ $H_{3}P{O}_4$ $(d=1.5\frac{g}{mL})$ by weight is :

• A. 6 M
• B. 15 M
• C. 10 M
• D. 4 M

Answer 1

The correct option is B 15 M

Taking the basis as 100 g of the solution, mass of the acid is 98 g since $\%w/w=98\%$.
The density of solution is $1.5gm{L}^{-1}$
Molar mass of the acid, $M_o=98gmo{l}^{-1}$
Molarity:
$=\frac{moles}{V_{solution}inL}=\frac{\%w/w\left(g\right)}{M_o\left(g/mol\right)}\times \frac{d\left(g/mL\right)}{100\left(g\right)}\times \frac{1000\left(mL\right)}{1\left(L\right)}=\frac{\%w/w}{M_o}\times d\times 10$
$M=\frac{98}{98}\times 1.5\times 10=15M$