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If $2 \cdot 0 \mathrm{~g}$ of $\mathrm{NaOH}$ is dissolved in $500 \mathrm{~cm}^{3}$ of water, what is molarity of solution?
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The correct answer is:
$0 \cdot 1 \mathrm{~mol} \mathrm{dm}^{-3}$
Molarity $=0.10 \cdot \mathrm{mol} \cdot \mathrm{L}^{-1}$ with respect to $\mathrm{NaOH}$
Explanation: Molarity $=\frac{\text { Moles of solute }}{\text { Volume of solution }}$
Molarity $=\frac{2 \cdot g}{40.0 \cdot g \cdot m o l^{-1}} \times \frac{1}{500 \cdot m L \times 10^{-3} \cdot L \cdot m L^{-1}}$
Molarity $=0.1 \mathrm{~mol} \mathrm{dm}^{\wedge}-3$
Explanation: Molarity $=\frac{\text { Moles of solute }}{\text { Volume of solution }}$
Molarity $=\frac{2 \cdot g}{40.0 \cdot g \cdot m o l^{-1}} \times \frac{1}{500 \cdot m L \times 10^{-3} \cdot L \cdot m L^{-1}}$
Molarity $=0.1 \mathrm{~mol} \mathrm{dm}^{\wedge}-3$
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