Two gases $A$ and $B$ having the same volume diffuse through a porous partition in $20$ and $10$ seconds respectively. The molecular mass of $A$ is $49 u$ . Molecular mass of $B$ will be:

50.00 u

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12.25 u

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6.50 u

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25.00 u

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Solution

The correct option is B $12.25 u$
We know that,
$\frac{r_{A}}{r_{B}} = \sqrt{\frac{M_{B}}{M_{A}}}$
We know that, $r = \frac{V}{t}$
Where, $r=rate V=Velocity t=time$

$\frac{t_{B}}{t_{A}} = \sqrt{\frac{M_{B}}{M_{A}}} \Rightarrow \frac{10}{20} = \sqrt{\frac{M_{B}}{49}} \Rightarrow {(\frac{10}{20})}^{2} = \frac{M_{B}}{49} \Rightarrow M_{B} = \frac{49 \times 100}{400} = 12.25 u$

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Two gases A and B having the same volume diffuse through a porous partition in 20 and 10 seconds respectively. The molecular mass of A is 49 u. Molecular mass of B will be:

The correct option is B 12.25 uWe know that, rArB=√MBMA We know that, r=Vt Where, r=rateV=Velocityt=time tBtA=√MBMA⇒1020=√MB49⇒(1020)2=MB49⇒MB=49×100400=12.25 u

Two gases $A$ and $B$ having the same volume diffuse through a porous partition in $20$ and $10$ seconds respectively. The molecular mass of $A$ is $49 u$ . Molecular mass of $B$ will be: