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Concentration Terms - An Introduction (w/w etc)

10 g of MnO...

Question

$10 g$ of $M n O_{2}$ on reaction with HCI forms $2.24 L$ of ${C I_{2}}_{(g)}$ at NTP, the percentage impurity of $M n O_{2}$ is:

$M n O_{2} + 4 H C I \to M n C I_{2} + C I_{2} + 2 H_{2} O$

87%

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25%

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33.3%

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13%

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Solution

The correct option is D 13%
Here, as per the given equation, $1$ mole of $C l_{2}$ is formed from $1$ mole of $M n O_{2}$

Therefore,
Number of moles of $C l_{2}$ is $n = \frac{2.24 L}{22.4 L} = 0.1 m o l$

Therefore, Mass of $C l_{2}$ is given by $W = 87 g / m o l \times 0.1 m o l = 8.7 g$

Impurity in $M n O_{2} = 10 g - 8.7 g = 1.3 g$

Percentage impurity $= \frac{1.3 g}{10 g} \times 100 = 13 %$

Hence, the correct option is $D$

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