Q51 / 180 · JIPMER 2019 · Chemistry

Calculate molarity of one litre solution of $22.2 \mathrm{gm} \mathrm{CaCl}_2$.

  • A
    0.4 M
  • Correct answer
    B
    0.2 M
  • C
    0.8 M
  • D
    0.6 M

Solution

$\begin{aligned} & \text { Molarity (M) } \\ & =\frac{\text { mass of solute (in } \mathrm{g})}{\text { molecular weight of solute }}=\frac{222 \mathrm{~g}}{111 \mathrm{~g} \mathrm{~mol}^{-1} \times 1 \mathrm{~L}} \\ & \times \text { Volume of solution (in } \mathrm{L} \text { ) } \\ & \qquad\left[\because \text { Molecular weight of } \mathrm{CaCl}_2=111 \mathrm{~g} \mathrm{~mol}^{-1}\right] \\ & =0.2 \mathrm{~mol} \mathrm{~L}^{-1}=0.2 \mathrm{M} .\end{aligned}$

Hence, the correct answer is option: 'B'.

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