JEE MAIN - Chemistry (2023 - 1st February Morning Shift - No. 21)
The density of $$3 \mathrm{M}$$ solution of $$\mathrm{NaCl}$$ is $$1.0 \mathrm{~g} \mathrm{~mL}^{-1}$$. Molality of the solution is ____________ $$\times 10^{-2} \mathrm{~m}$$. (Nearest integer).
Given: Molar mass of $$\mathrm{Na}$$ and $$\mathrm{Cl}$$ is $$23$$ and $$35.5 \mathrm{~g} \mathrm{~mol}^{-1}$$ respectively.
Answer
364
Explanation
$\mathrm{m}=\frac{1000 \mathrm{M}}{1000 \rho-\mathrm{M} . \mathrm{m_w}}=\frac{1000 \times 3}{1000-3 \times(58.5)}$ $=\frac{3000}{(1000-175.5)}=3.638$
$=363.8 \times 10^{-2}$
Nearest integer $=364$
$=363.8 \times 10^{-2}$
Nearest integer $=364$
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