Verified Solution • 2025–26 Session moderate

QJEEM23Chb1ff: JEE Main 2023 Chemistry - Class 11 JEEM23Chb1ff Solution: The molality of a solution of di-bromine solution in (carbon…

The molality of a 10%(v/v)10 \%(\mathrm{v} / \mathrm{v}) solution of di-bromine solution in CCl4\mathrm{CCl}_{4} (carbon tetrachloride) is ' xx '. x=x= ____________ ×10−2 M\times 10^{-2} ~\mathrm{M} . (Nearest integer)

[Given : molar mass of Br2=160 g mol−1\mathrm{Br}_{2}=160 \mathrm{~g} \mathrm{~mol}^{-1}

atomic mass of C=12 g mol−1\mathrm{C}=12 \mathrm{~g} \mathrm{~mol}^{-1}

atomic mass of Cl=35.5 g mol−1\mathrm{Cl}=35.5 \mathrm{~g} \mathrm{~mol}^{-1}

density of dibromine =3.2 g cm−3=3.2 \mathrm{~g} \mathrm{~cm}^{-3}

density of CCl4=1.6 g cm−3\mathrm{CCl}_{4}=1.6 \mathrm{~g} \mathrm{~cm}^{-3} ]

Step-by-Step Verified Solution

Mass of 10 mL10 \mathrm{~mL} of Br2=10×3.2=32gm\mathrm{Br}_2=10 \times 3.2=32 \mathrm{gm}

Mass of 90 mLLCCl4=90×1.6=144gm90 \mathrm{~mL}^{\mathrm{L}} \mathrm{CCl}_4=90 \times 1.6=144 \mathrm{gm}

Molality of Br2\mathrm{Br}_2 solution in CCl4\mathrm{CCl}_4

amp;=32×1000160×144amp;=1.39Mamp;=139×10−2\begin{aligned}& =\frac{32 \times 1000}{160 \times 144} \\\\& =1.39 \mathrm{M} \\\\& =139 \times 10^{-2}\end{aligned}

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