Question

6. 50 ml of a gas A diffused through a membrane in the same time as for the diffusion of 30 ml of a gas B under identical pressure temperature conditions. If molecular weight of A=81, that of B would be _________

Answer 1

use the formula pv=nRt
v=mass/density
pw=dRt ___2
find density of second gas
it may be same for first gas
substitute in eq 2
you will get the answer

Answer 2

Answer : The molecular weight of gas B will be, 3.77 g/mole

Solution : Given,

Volume of gas A = 6.5 ml

Volume of gas B = 30 ml

Molecular weight of gas A = 81 g/mole

Rate of effusion : It is defined as the volume of gas effused in a given time 't'.

Formula used : $Rate=\frac{Volume}{Time}$

Or,

Rate of effusion : It is defined as the rate of effusion is directly proportional to the square root of the mass of the gas.

$\text{ Rate of effusion}\propto \frac{1}{\sqrt{\text{ Mass of gas}}}$

From the two expressions, we conclude that the relation between the volume and the molecular mass of gas in the same time is,

$V\propto \frac{1}{\sqrt{M}}$

or, $\sqrt{\frac{M_B}{M_A}}=\frac{V_A}{V_B}$     .........(1)

where,

$M_A$ = molecular mass of gas  A

$M_B$ = molecular mass of gas  B

$V_A$ = volume of gas  A

$V_B$ = volume of gas  B

Now put all the given values in the expression (1), we get

$\sqrt{\frac{M_B}{81g/mole}}=\frac{6.5ml}{30ml}$

$M_B=3.77g/mole$

Therefore, the molecular weight of gas B will be, 3.77 g/mole