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Molarity

6.13 g sample...

Question

$6.13$ g sample of $K_{2} C r_{2} O_{7}$ ( $96 %$ purity) is allowed to react with $320$ mL of $H C l$ solution having density $1.15$ g/mL and containing $30 %$ by mass of $H C l$ . What mass of $C l_{2}$ is liberated?
$K_{2} C r_{2} O_{7} + 14 H C l ⟶ 2 K C l + 3 C r C l_{3} + 7 H_{2} O + 3 C l_{2}$

4.7

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4.3

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7.3

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None of these

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Solution

The correct option is A $4.3$ g
$m_{H C l} = \frac{30}{36.5 \times \frac{100}{1.15 \times 1000}} = 9.45 M$
Millimoles of $H C l = 9.45 \times 320 = 3.24$
Millimoles of $K_{2} C r_{2} O_{7} = 6.13 \times \frac{96}{100} \times \frac{1000}{294} = 20.02$
$∴$ Millimoles of $C l_{2}$ formed $= 20.03 \times 3$
$∴ w_{C l_{2}} = \frac{20.03 \times 3 \times 71}{1000} = 4.26$ g
Hence, mass of $C l_{2}$ liberated is $4.26$ g.

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