Some Basic Concepts of Chemistry 1 Question 16
37. $3.0 \mathrm{~g}$ of a salt of molecular weight 30 is dissolved in $250 \mathrm{~g}$ water. The molarity of the solution is
(1983, 1M)
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Solution:
- Molarity $=\frac{\text { Number of moles of solute }}{\text { Volume of solution in litre }}$
$$ \begin{aligned} & =\frac{\text { Weight of solute }}{\text { Molar mass }} \times \frac{1000}{\text { Volume in } \mathrm{mL}} \ & =\frac{3}{30} \times \frac{1000}{250}=0.4 \mathrm{M} \end{aligned} $$
