(a) $w = 1000\ g\left( H_{2}O \right);n = 1\ mol$ $N = \frac{W}{M} = \frac{1000}{18} = 55.55$ $\mathcal{X}_{solute} = \frac{n}{n + N} + \frac{1}{1 + 55.55}$ $= 0.018$
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