Question

The molarity of $98\%$ $H_{2}S{O}_4$ $(d=1.8\frac{g}{ml})$ by weight is :

• A. $6M$
• B. $18M$
• C. $10M$
• D. $4M$

Answer 1

The correct option is B $18M$

Taking the basis as $100\text{g}$ of the solution, mass of the acid is $98\text{g}$ since $\%w/w=98\%$.
The density of solution is $1.8{\text{g mL}}^{-1}$
Molar mass of the acid, $M_o=98{\text{g mol}}^{-1}$
Molarity:
$=\frac{moles}{V_{solution}inL}=\frac{\%w/w\left(g\right)}{M_o\left(g/mol\right)}\times \frac{d\left(g/mL\right)}{100\left(g\right)}\times \frac{1000\left(mL\right)}{1\left(L\right)}=\frac{\%w/w}{M_o}\times d\times 10$
$M=\frac{98}{98}\times 1.8\times 10=18M$