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What is the weight (in gram) of $\mathrm{Na}_2 \mathrm{CO}_3$ (molar mass $=106)$ present in $250 \mathrm{~mL}$ of its $0.2 \mathrm{M}$ solution?
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The correct answer is:
5.3
Molarity
$$
\begin{aligned}
& =\frac{\text { Weight of solute }}{\text { Molecular weight of solute }} \times \frac{1000}{\text { volume }(\text { in mL })} \\
& 0.2=\frac{\text { Weight of solute }}{106} \times \frac{1000}{250} \\
& \therefore \quad \text { Weight of solute }=\frac{0.2 \times 106 \times 250}{1000}=5.3 \mathrm{gm}
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{\text { Weight of solute }}{\text { Molecular weight of solute }} \times \frac{1000}{\text { volume }(\text { in mL })} \\
& 0.2=\frac{\text { Weight of solute }}{106} \times \frac{1000}{250} \\
& \therefore \quad \text { Weight of solute }=\frac{0.2 \times 106 \times 250}{1000}=5.3 \mathrm{gm}
\end{aligned}
$$
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