JEE MAIN - Chemistry (2023 - 1st February Evening Shift - No. 21)
The molality of a $$10 \%(\mathrm{v} / \mathrm{v})$$ solution of di-bromine solution in $$\mathrm{CCl}_{4}$$ (carbon tetrachloride) is '$$x$$'. $$x=$$ ____________ $$\times 10^{-2} ~\mathrm{M}$$. (Nearest integer)
[Given : molar mass of $$\mathrm{Br}_{2}=160 \mathrm{~g} \mathrm{~mol}^{-1}$$
atomic mass of $$\mathrm{C}=12 \mathrm{~g} \mathrm{~mol}^{-1}$$
atomic mass of $$\mathrm{Cl}=35.5 \mathrm{~g} \mathrm{~mol}^{-1}$$
density of dibromine $$=3.2 \mathrm{~g} \mathrm{~cm}^{-3}$$
density of $$\mathrm{CCl}_{4}=1.6 \mathrm{~g} \mathrm{~cm}^{-3}$$]
Answer
139
Explanation
Mass of $10 \mathrm{~mL}$ of $\mathrm{Br}_2=10 \times 3.2=32 \mathrm{gm}$
Mass of $90 \mathrm{~mL}^{\mathrm{L}} \mathrm{CCl}_4=90 \times 1.6=144 \mathrm{gm}$
Molality of $\mathrm{Br}_2$ solution in $\mathrm{CCl}_4$
$$ \begin{aligned} & =\frac{32 \times 1000}{160 \times 144} \\\\ & =1.39 \mathrm{M} \\\\ & =139 \times 10^{-2} \end{aligned} $$
Mass of $90 \mathrm{~mL}^{\mathrm{L}} \mathrm{CCl}_4=90 \times 1.6=144 \mathrm{gm}$
Molality of $\mathrm{Br}_2$ solution in $\mathrm{CCl}_4$
$$ \begin{aligned} & =\frac{32 \times 1000}{160 \times 144} \\\\ & =1.39 \mathrm{M} \\\\ & =139 \times 10^{-2} \end{aligned} $$
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