Question

What is the molarity of $H_{2}S{O}_4$ solution that has a density of $1.84gm/cc$ and contains $98\%$ by mass of $H_{2}S{O}_4$?

[Given atomic mass of $S=32$]

• A. $4.18M$
• B. $8.14M$
• C. $18.4M$
• D. $18M$

Answer 1

The correct option is C $18.4M$

Molar mass of $H_{2}S{O}_4$ $=98gm/mol=Mm$

$D=\text{density}=1.84gm/cc$

$w/w=98\%$

$V=1d{m}^3=1L=1000ml=1000cc$

Mass of solution $=V\times D$

$M=1000\times 1.84=1840gmper1000cc$

Using w/v, $98\%of1840gm=H_{2}S{O}_4=\frac{98}{100}\times 1840=1803.2gm=m$

No. of moles $=\frac{m}{Mm}=\frac{1803.2}{98}=18.4molesper1000cc$

Thus, molarity $=18.4M\left(18.4mol/d{m}^3\right)$

Hence, the correct option is $C$