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The amount of sugar $(C_{12} H_{22} O_{11})$ required to prepare 2L of its 0.1 M aqueous solution is

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17.1 g

68.4 g

136.8 g

34.2 g

answer is A.

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Detailed Solution

Molarity $= \frac{(n)_{solute}}{V_{solution} (in lit)}$

$\begin{array}{l} 0 .1 = \frac{wt . / 342}{2} \\ wt (C_{12} H_{22} O_{11}) = 68 .4 gram \end{array}$

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