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Active mass of 1 gram of hydrogen at STP is X. Active mass of 10 grams of limestone in 10 litre vessel is Y. Y : X is

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$1 : 2$

$22.4 : 1$

$1 : 1$

$4 : 1$

answer is B.

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Detailed Solution

$n_{C a C O_{3}} = \frac{10}{100} = 0.1$ =

For solid $C a C O_{3}$ , active mass is one.
Number of moles of $H_{2} = \frac{1}{2} = 0.5$
Active mass $= \frac{0.5}{11.2} = \frac{1}{22.4}$
$X : Y = \frac{1}{22.4} : 1 = 1 : 22.4$

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