Some Basic Concepts of Chemistry - Result Question 71
####9. The amount of $\operatorname{sugar}\left(C _{12} H _{22} O _{11}\right)$ required to prepare $2 L$ of its $0.1 M$ aqueous solution is
(2019 Main, 10 Jan II)
(a) $17.1 g$
(b) $68.4 g$
(c) $136.8 g$
(d) $34.2 g$
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Solution:
- Molarity $=\frac{\text { Number of moles of solute }(n)}{\text { Volume of solution (in L) }}$
Also, $\quad n=\frac{w _B(g)}{M _B\left(g mol^{-1}\right)}$
$\therefore \quad$ Molarity $=\frac{w _B / M _B}{V}$
Given, $w _B=$ mass of solute $(B)$ in $g$
$$ M _B=\text { Gram molar mass of } B\left(C _{12} H _{22} O _{11}\right)=342 g mol^{-1} $$
Molarity $=0.1 M$
Volume $(V)=2 L$
$\Rightarrow \quad 0.1=\frac{w _B / 342}{2} \Rightarrow w _B=0.1 \times 342 \times 2 g=68.4 g$
