A solution of acetic acid has molarity equal to $1.35 \mathrm{M}$ and molality equal to $1.45 \mathrm{~mol}…

A solution of acetic acid has molarity equal to $1.35 \mathrm{M}$ and molality equal to $1.45 \mathrm{~mol} \mathrm{~kg}^{-1}$. The density of solution will be
  1. $1.251 \mathrm{~g} \mathrm{~mL}^{-1}$
  2. $1.125 \mathrm{~g} \mathrm{~mL}^{-1}$
  3. $1.012 \mathrm{~g} \mathrm{~mL}^{-1}$
  4. $0.994 \mathrm{~g} \mathrm{~mL}^{-1}$

Solution

For $1.35 \mathrm{M}$ solution, we will have $\quad n_{2}=1.35 \mathrm{~mol} \quad$ and $\quad V=1000 \mathrm{~mL}$ The solution is also $1.45 \mathrm{~mol} \mathrm{~kg}^{-1}$. The mass of solvent to have $1.35 \mathrm{~mol}$ of solute will be
$m_{1}=\left(\frac{1.35 \mathrm{~mol}}{1.45 \mathrm{~mol} \mathrm{~kg}^{-1}}ight)=0.93103 \mathrm{~kg}=931.03 \mathrm{~g}$
The mass of solution will be $m=m_{1}+n_{2} M_{2}=931.03 \mathrm{~g}+(1.35 \mathrm{~mol})\left(60 \mathrm{~g} \mathrm{~mol}^{-1}ight)=1012.03 \mathrm{~g}$ The density of the solution will be $\quad ho=\frac{m}{V}=\frac{1012.03 \mathrm{~g}}{1000 \mathrm{~mL}}=1.012 \mathrm{~g} \mathrm{~mL}^{-1}$ .

Asked in: JEE-TOPICTESTS-CHEMISTRY

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