Question

The molality of $15\%(wt./vol.)$ solution of $H_{2}S{O}_4$ of density $1.1g/c{m}^3$ is approximately.

• A. $1.2$
• B. $1.4$
• C. $1.8$
• D. $1.6$

Answer 1

The correct option is D $1.6$

The solution have 15% (mass/volume)

let mass of solute = 15g

volume of solution = 100 ml

we have, density = $1.1g/c{m}^3$

so, mass of solution = $density\ast volume=1.1\times 100=110g$
$\text{mass of solvent(g)=}110-15g$

molecular mass of $H_{2}S{O}_4=98$

so, moles of $H_{2}S{O}_4$ = $\frac{\text{given weight}}{\text{molecular weight}}=\frac{15}{98}=0.153moles$

molality = $\frac{\text{moles of solute}}{\text{mass of solvent(g)}}\times 1000$

so, molality =$\frac{0.153}{95}\times 1000=1.6$