Question

127 ml of a certain gas diffuse in the same time as 100 ml of chlorine under the same conditions. Calculate
the value of xif molecular mass of unknown gas is expressed as x x 11 amu​

Answer 1

it is based on Graham's law of diffusion,

a/c to law of diffusion, rate of diffusion or effusion is inversely proportional to square root of molecular weight.

$\frac{V_1t_2}{V_2t_1}=\sqrt{\frac{M_2}{M_1}}$

for , $t_1=t_2$

$\frac{V_1}{V_2}=\sqrt{\frac{M_2}{M_1}}$

$V_1=127ml,V_2=100ml$

$M_1=M$ and $M_2$ = 71g g/mol

now, 100/127 = √{M/71}

taking square both sides,

or, (100)²/(127)² = M/71

or, M/71 = (100/127)²

or, M/71 = (1/1.27)²

or, M = 71/(1.27)²

or, M = 71/(1.6129)

hence, M = 44.020088 ≈ 44

given molecular mass of gas = 11 × x

or, 44 = 11x

or, x = 44/11 = 4

hence, value of x = 4.

Answer 2

Using the Grams law of diffusion,

$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1} }$

M₂ is the molar mass of chlorine, and M₁ is the molar mass of Gas X.

Now, Time is same, thus,

 $\frac{V_1}{V_2} = \sqrt{\frac{M_2}{M_1} }$

Where, V₁ and V₂ is the amount of volume of unknown and Chlorine gas diffused respectively.

  ∴ 127/100 = √(71/M)

      71/M = 1.27²

∴ M = 71/1.27²

∴ M =  44.02 gm.

Now, for value of x,

 x = Molar mass/11

x = 44.02/11

x ≈ 4

Hope it helps.