Question

360cm3 methane gas diffused through a porous membrane in 15min . under similar conditions 120cm3 another gas diffused in 10min f. Find the molar mass of second gas.

Answer 1

Explanation:

Rate of different r=

f

V

r=

M

1

therefore,

r

g

r

CH

4

=

M

CH

4

M

g

=

V

g

×t

CH

4

V

CH

4

×t

g

16

M

g

=

120×15

360×10

16

M

g

=2

M

g

=64.