MCQ
$25\, g$ of a solute of molar mass $250\, g\, mol^{-1}$ is dissolved in $100\, ml$ of water to obtain a solution whose density is $1.25\, g\, ml^{-1}$. The molarity and molality of the solution are respectively
  • A
    $0.75$ and $1$
  • B
    $0.8$ and $1$
  • C
    $1$ and $0.8$
  • $1$ and $1$

Answer

Correct option: D.
$1$ and $1$
d
No. of moles of solute $=\frac{25}{250}=0.1$ moles Molarity $=\frac{\text { No. of moles of solute }}{\text { Volume of solvent in litre }}=\frac{0.1}{\frac{100}{1000}}=1 M$

Moles of solute $=\frac{25 \mathrm{g}}{250 \mathrm{g} / \mathrm{mol}}=0.1 \mathrm{mol}$

Density of water $=1 \mathrm{g} / \mathrm{ml}$

Mass of solvent (water) $=100\; ml \times 1 \frac{\mathrm{g}}{\mathrm{ml}}=100 \mathrm{g}=0.1 \mathrm{kg}$

Molality $=\frac{\text { Moles of solute }}{\text { Mass of solvent (in kg) }}$

Molality $=\frac{0.1 \mathrm{mol}}{0.1 \mathrm{kg}}=1 \mathrm{m}$

Molality of solution is $1 \mathrm{m}$

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