Some Basic Concepts of Chemistry 1 Question 9
9. The amount of $\operatorname{sugar}\left(\mathrm{C}{12} \mathrm{H}{22} \mathrm{O}_{11}\right)$ required to prepare $2 \mathrm{~L}$ of its $0.1 \mathrm{M}$ aqueous solution is
(2019 Main, 10 Jan II)
(a) $17.1 \mathrm{~g}$
(b) $68.4 \mathrm{~g}$
(c) $136.8 \mathrm{~g}$
(d) $34.2 \mathrm{~g}$
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Solution:
- Molarity $=\frac{\text { Number of moles of solute }(n)}{\text { Volume of solution (in L) }}$
Also, $\quad n=\frac{w_{B}(\mathrm{~g})}{M_{B}\left(\mathrm{~g} \mathrm{~mol}^{-1}\right)}$
$\therefore \quad$ Molarity $=\frac{w_{B} / M_{B}}{V}$
Given, $w_{B}=$ mass of solute $(B)$ in $\mathrm{g}$
$$ M_{B}=\text { Gram molar mass of } B\left(\mathrm{C}{12} \mathrm{H}{22} \mathrm{O}_{11}\right)=342 \mathrm{~g} \mathrm{~mol}^{-1} $$
Molarity $=0.1 \mathrm{M}$
Volume $(V)=2 \mathrm{~L}$
$\Rightarrow \quad 0.1=\frac{w_{B} / 342}{2} \Rightarrow w_{B}=0.1 \times 342 \times 2 \mathrm{~g}=68.4 \mathrm{~g}$
