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JIPMER 2019
Exam
JIPMER
Year
2019
Subject
Chemistry
Chapter
Solutions

JIPMER 2019 · Chemistry · Solutions

Calculate molarity of one litre solution of 22.2 gm CaCl_2 .

Calculate molarity of one litre solution of 22.2gmCaCl222.2 \mathrm{gm} \mathrm{CaCl}_2.

Options

  1. A0.4 M
  2. B0.2 M
  3. C0.8 M
  4. D0.6 M

Correct answer

B. 0.2 M

Step-by-step solution

 Molarity (M) = mass of  solute (in g) molecular weight of solute =222g111gmol−1×1L× Volume of solution (in L ) [∵ Molecular weight of CaCl2=111gmol−1]=0.2molL−1=0.2M.\begin{aligned} & \text{ Molarity (M) } \\ & =\frac{\text{ mass\text{ 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}
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