a gas with the molecular formula CnH2n+2 diffuse through a porous plug

1.	a gas with the molecular formula CnH2n+2 diffuse through a porous plug @ one sixth of the rate of diffusion of hydrogen under similar conditions find the value of n
Answer» Answer: The value of N is 5. Explanation: RATE of diffusion of hydrogen gas = r Rate of diffusion of unknown gas = R = $\frac{1}{<klux>6</klux>}r$ r = 6 × R Molar mass of hydrogen gas = 2 g/mol Molar mass of unknown gas = M Graham's Law STATES that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation: $\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$ So, $\frac{r}{R}=\sqrt{\frac{M}{2 g/mol}}$ $\frac{6R}{R}=\sqrt{\frac{M}{2}}$ $M=\frac{36\times R^2\times 2 g/mol}{R^2}=72 g/mol$ Molar mass of unknown gas, $C_nH_{2n+2}= M = 72g/mol$ $n\times 12 g/mol+1 g/mol(2n+2)=72 g/mol$ n = 5 The value of n is 5.

a gas with the molecular formula CnH2n+2 diffuse through a porous plug @ one sixth of the rate of diffusion of hydrogen under similar conditions find the value of n

Answer»

Answer:

The value of N is 5.

Explanation:

RATE of diffusion of hydrogen gas = r

Rate of diffusion of unknown gas = R = $\frac{1}{<klux>6</klux>}r$

r = 6 × R

Molar mass of hydrogen gas = 2 g/mol

Molar mass of unknown gas = M

Graham's Law STATES that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$