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 $1.5{\text{g mL}}^{-1}$
Molar mass of the acid, $M_o=98{\text{g mol}}^{-1}$
Molarity:
$=\frac{moles}{V_{solution}\text{in L}}=\frac{\%w/w\left(\text{g}\right)}{M_o\left(\text{g/mol}\right)}\times \frac{d\text{(g/mL)}}{100\text{(g)}}\times \frac{1000\text{(mL)}}{1\text{(L)}}=\frac{\%w/w}{M_o}\times d\times 10$
$M=\frac{98}{98}\times 1.5\times 10=15\text{M}$