A 0.50 g sample containing only anhydrous ${FeCl}_{3}$ ( molar mass: $162.5 g {mol}^{- 1}$ ) and ${AlCl}_{3}$ (molar mass: $133.5 g$ ${mol}^{- 1}$ yielded $1.435 g$ of $A g Cl$ (molar mass: $143.5 g {mol}^{- 1}$ ). The mass of ${FeCl}_{3}$ in the sample is

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0.3126 g

0.2345 g

0.4157 g

0.1567 g

answer is A.

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Detailed Solution

Amount of AgCl obtained, $n = \frac{m}{M} = \frac{1 .435 g}{143 .5 {gmol}^{- 1}} = 0 .01 mol$

Let x be the mass of ${FeCl}_{3}$ , then

Amount of ${FeCl}_{3}$ $= \frac{x}{162 g {mol}^{- 1}}$ and Amount of ${AlCl}_{3} = \frac{0.50 g - x}{133.5 g {mol}^{- 1}}$

Hence, $3 [\frac{x}{162 g {mol}^{- 1}} + \frac{0.50 g - x}{133.5 g {mol}^{- 1}}] = 0.01 mol$

Solving for x, we get

$x = \frac{1}{(28 .5)} [\frac{0 .01 \times 162 \times 133 .5}{3} - 81] g = 0 .3126 g$

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