At the same temperature and pressure,the rate | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) According to Graham's Law of diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $r \propto \f

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) According to Graham's Law of diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $r \propto \frac{1}{\sqrt{M}}$.Therefore,$\frac{r_X}{r_{	ext{gas}}} = \sqrt{\frac{M_{	ext{gas}}}{M_X}}$.Given that $r_X = 3 \sqrt{3} 	imes r_{	ext{gas}}$,we have $\frac{r_X}{r_{	ext{gas}}} = 3 \sqrt{3}$.Substituting the values: $3 \sqrt{3} = \sqrt{\frac{54}{M_X}}$.Squaring both sides: $(3 \sqrt{3})^2 = \frac{54}{M_X}$.$27 = \frac{54}{M_X}$.$M_X = \frac{54}{27} = 2 \ g \ mol^{-1}$.

At the same temperature and pressure,the rate of diffusion of gas $X$ is $3 \sqrt{3}$ times that of a gaseous hydrocarbon of molar mass $54 \ g \ mol^{-1}$. The molar mass of $X$ in $g \ mol^{-1}$ is

Similar Questions

On what factors does the stability of the state of matter depend?

The rate of diffusion of a gas $A$ is $\sqrt{5}$ times more than that of gas $B$. If the molar mass of $A$ is $x \ g \ mol^{-1}$, the molar mass of $B \ (\text{in } g \ mol^{-1})$ is (in $x$)

What is the difference between a barometer and a manometer?

$A$ sample of $N_2$ gas at $0.8 \, atm$ pressure takes $38 \, s$ to effuse through a pinhole. $A$ sample of $1 \, mol$ of an unknown gas '$X$' at $1.6 \, atm$ pressure takes $57 \, s$ to effuse through the same pinhole. The molar mass of gas '$X$' in $g/mol$ is .............

When the temperature is raised,the viscosity of a liquid decreases. This is because:

Vedclass Test Series

Exam Paper Generator

Online Exam Module