Question

If 280cm3 of hydrogen diffuses in 40secs, how long will it take for 490cm3 of a gas x, whose vapor density is 25 to diffuse under the same condition (RMM hydrogen =2)​

Answer 1

The time taken by gas x : 350 s

Further explanation

Graham's law: the rate of effusion of a gas is inversely proportional to the square root of its molar masses or

the effusion rates of two gases = the square root of the inverse of their molar masses:

`\rm (r_1)/(r_2)=\sqrt{(M_2)/(M_1) }`

Molar mass = 2 x vapour density(VD)

r₁=rate Hydrogen = 280 cm³ / 40 s = 7 cm³/s

M₁=molar mass Hydrogen=2 g/mol

M₂=molar mass gas x=

`\tt 2* VD=2* 25=50~g/mol`

`\tt (7)/(r_2)=\sqrt{(50)/(2) }\\\\(7)/(r_2)=5\\\\r_2=1.4~cm^3/s`

`\tt r_2=(V_2)/(t_2)\\\\t_2=(V_2)/(r_2)\\\\t_2=(490~cm^3)/(1.4~cm^3/s)\\\\t_2=350~s`