MCQ
Molarity $(M)$ of an aqueous solution containing $\mathrm{x} g$ of anhyd. $\mathrm{CuSO}_4$ in $500 \mathrm{~mL}$ solution at $32^{\circ} \mathrm{C}$ is $2 \times 10^{-1} \mathrm{M}$. Its molality will be. . . . . .$\times 10^{-3} \mathrm{~m}$ (nearest integer). [Given density of the solution $=1.25 \mathrm{~g} / \mathrm{mL}$.]
  • A
    $160$
  • $164$
  • C
    $167$
  • D
    $168$

Answer

Correct option: B.
$164$
b
$\mathrm{M}_{\mathrm{rol}^{\mathrm{n}}}=\mathrm{v}_{\mathrm{sol}^{\mathrm{n}}} \times \mathrm{d}_{\mathrm{sol}^{\mathrm{n}}}$

$=500 \times 1.25=625 \mathrm{~g}$

$\text { Mass of solute }(x)=0.2 \times 0.5 \times 159.5$

$=15.95$

$\mathrm{n}_{\text {solute }}=0.1 \text {, }$

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

$=625-15.95$

$=609.05$

$\mathrm{~m}=\frac{0.1}{\frac{609.05}{1000}}$

$\mathrm{~m}=0.164=164 \times 10^{-3}$

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