Question

Molarity and molality of a solution of a liquid (mol wt=50) in aqueous solution is $9$ and $10$ respectively. What is the density of solution?

• A. $1g/cc$
• B. $0.95g/cc$
• C. $1.05g/cc$
• D. $1.35g/cc$

Answer 1

The correct option is D $1.35g/cc$

$\text{Molarity}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}}\times \frac{1000}{\text{Volume of solution}}$

$M=\frac{w}{m^{\prime }}\times \frac{1000}{v}⟹w=\frac{M\times m^{\prime }\times v}{1000}⟶\left(1\right)$

$\text{Molality}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}}\times \frac{1000}{\text{Mass of solvent}}$

$w=\frac{w}{m^{\prime }}\times \frac{1000}{w^{\prime }}⟹w=\frac{m\times m^{\prime }\times w^{\prime }}{1000}⟶\left(2\right)$

$\text{Mass of solution}=\text{Mass of solvent}+\text{Mass of solute}$

$w"=w+w^{\prime }$

$vd=w+w^{\prime }$

$w^{\prime }=vd-w⟶\left(3\right)$

Putting value of $w^{\prime }$ in equation $\left(2\right)$ and on equate equation $\left(1\right)$ and $\left(2\right)$,

we get,

$\frac{M\times m^{\prime }\times v}{1000}=\frac{m\times m^{\prime }\times (vd-w)}{1000}$

$M\times v=mvd-mw$

$M\times v=mvd-m\left(\frac{M{m}^{\prime }v}{1000}\right)$

$M=m(d-\frac{m^{\prime }M}{100})$

OR,

$d=M(\frac{1}{m}+\frac{m^{\prime }}{1000})$

$\text{density}=\text{molarity}(\frac{1}{\text{molality}}+\frac{\text{molecular weight}}{1000})$

$d=9(\frac{1}{10}+\frac{50}{1000})=1.35g/cc$