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:  

$\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }$

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=

$\tt 2\times VD=2\times 25=50~g/mol$

$\tt \dfrac{7}{r_2}=\sqrt{\dfrac{50}{2} }\\\\\dfrac{7}{r_2}=5\\\\r_2=1.4~cm^3/s$

$\tt r_2=\dfrac{V_2}{t_2}\\\\t_2=\dfrac{V_2}{r_2}\\\\t_2=\dfrac{490~cm^3}{1.4~cm^3/s}\\\\t_2=350~s$