At $546^{o} C$ and $1$ $a t m$ pressure, $0.062$ grams of gas occupies a volume of $33.6$ $m l$ . The molecular weight of the gas is:

62

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124

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67.2

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31

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Solution

The correct option is D $124$
Given,
$P = 1 a t m, w = 0.062 g, T = 546^{o} C = 819 K, V = 33.6 m l = 33.6 \times 10^{- 3} L$

$M = \frac{W R T}{P V}$
$M = \frac{0.062 \times 0.082 \times 819}{33.6 \times 10^{- 3}} = 124$ .

Hence, the molecular weight of the gas is $124.$
Option B is correct.

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At 546oC and 1 atm pressure, 0.062 grams of gas occupies a volume of 33.6 ml. The molecular weight of the gas is:

The correct option is D 124Given, P=1atm,w=0.062g,T=546oC=819K,V=33.6 ml=33.6×10−3LM=WRTPVM=0.062×0.082×81933.6×10−3=124. Hence, the molecular weight of the gas is 124.Option B is correct.

At $546^{o} C$ and $1$ $a t m$ pressure, $0.062$ grams of gas occupies a volume of $33.6$ $m l$ . The molecular weight of the gas is:

The correct option is D $124$
Given,
$P = 1 a t m, w = 0.062 g, T = 546^{o} C = 819 K, V = 33.6 m l = 33.6 \times 10^{- 3} L$

$M = \frac{W R T}{P V}$
$M = \frac{0.062 \times 0.082 \times 819}{33.6 \times 10^{- 3}} = 124$ .

Hence, the molecular weight of the gas is $124.$
Option B is correct.