80 mL of O_2 takes 2 minute to pass through | 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 given by $r = \frac{V}{t} \propto \frac{1}{\sqrt{M}}$.For two gases,the ratio

Vedclass · Accepted Answer

$\frac{120}{\sqrt{2}}$

Vedclass · Answer

(B) According to Graham's Law of Diffusion,the rate of diffusion $r$ is given by $r = \frac{V}{t} \propto \frac{1}{\sqrt{M}}$.For two gases,the ratio is $\frac{V_1 / t_1}{V_2 / t_2} = \sqrt{\frac{M_2}{M_1}}$.Given: $V_1 = 80 \ mL$,$t_1 = 2 \ min$,$M_1 (O_2) = 32 \ g/mol$,$t_2 = 3 \ min$,$M_2 (SO_2) = 64 \ g/mol$.Substituting the values: $\frac{80 / 2}{V_2 / 3} = \sqrt{\frac{64}{32}}$.$40 	imes \frac{3}{V_2} = \sqrt{2}$.$V_2 = \frac{120}{\sqrt{2}} \ mL$.

$80 \ mL$ of $O_2$ takes $2 \ minute$ to pass through a hole. What volume of $SO_2$ will pass through the same hole in $3 \ minute$?

Similar Questions

Which of the following graphs represents the behavior of one mole of an ideal gas at $1 \ atm$ pressure for volume $V$ versus temperature $T$?

Which of the following sets consists of gases with the same rate of diffusion?

$A$ mixture of $He$ and $CH_4$ in a molar ratio of $4:1$ is present in a container at a pressure of $20 \ bar$. Due to a hole in the container,the gas mixture leaks out. What will be the composition of the gas mixture effusing out initially (in $: 1$)?

Which of the following volume $(V)$-temperature $(T)$ plots represents the behaviour of one mole of an ideal gas at one atmospheric pressure?

Equal volumes of two monoatomic gases,$A$ and $B$ at same temperature and pressure are mixed. The ratio of specific heats $(C_P/C_V)$ of the mixture will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module