12 cm 3 of SO 2 ( g ) diffused through a porous membrane in 1 minute. Under similar conditions 120 cm 3 of another gas diffused in 5 minutes. The molar mass of the gas in g mol − 1 is

Question

12 cm 3 of SO 2 ​ ( g ) diffused through a porous membrane in 1 minute. Under similar conditions 120 cm 3 of another gas diffused in 5 minutes. The molar mass of the gas in g mol − 1 is, 32, 18, 44, 16

MARKS App · Accepted Answer

According to law of diffusion rate ( r ) α M ​ 1 ​ (molar mass) and r (rate) α time ( t ) volume ( V ) ​ i.e. r ( z ) ​ r ( SO 2 ​ ) ​ ​ = ( V / t ) ( z ) ​ ( V / t ) ( SO 2 ​ ) ​ ​ = ( l / M ​ ) C z ​ ​ [ l / M ​ ] ( SO 2 ​ ) ​ ​ or, V ( z ) ​ × t ( SO 2 ​ ) ​ V ( SO 2 ​ ) ​ × t ( z ) ​ ​ = M ( SO 2 ​ ) ​ M ( z ) ​ ​ ​ Given, $ V ( SO 2 ​ ) ​ t ( SO 2 ​ ) ​ V ( z ) ​ t ( z ) ​ ​ = 12 cm 3 = 1 minute = 1 × 60 = 60 sec = 120 cm 3 = 5 minutes = 5 × 60 = 300 sec ​ an d m o l a r ma ss (M) o f \mathrm{SO}_2=32+(2 	imes 16)=64 T h u s , \frac{12 	imes 300}{120 	imes 60}=\sqrt{\frac{M_z}{64}} or , \frac{1}{4}=\frac{M_z}{4} 	herefore \quad M_z=\frac{64}{4}=16 He n ce , m o l a r ma sso f g i v e n g a s i s =16 \mathrm{~g} \mathrm{~mol}^{-1}$

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