1.

If a gas diffuses at the rate of one-half as fast as O_2, find the molecular mass of the gas.

Answer»

Solution :APPLYING Graham.s law of diffusion .
`(r_(1))/(r_(2)) = sqrt((M_(2))/(M_(1))) , (1/2)/(1) = sqrt((32)/(M_(1)))`
SQUARING both sides of the equation
`((1)/(2))^(2) = (32)/(M_(1)) (or) (1)/(4) = (32)/(M_1)`
`M_1 = 128`
Thus the MOLECULAR mass of the UNKNOWN gas is 128.


Discussion

No Comment Found