Equal moles of hydrogen and oxygen gases are | vedclass

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

Question

(C) According to Graham's Law of Effusion,the rate of effusion $r$ is inversely proportional to the square root of the molar mass $M$: $r \propto \fra

Vedclass · Accepted Answer

$1/8$

Vedclass · Answer

(C) According to Graham's Law of Effusion,the rate of effusion $r$ is inversely proportional to the square root of the molar mass $M$: $r \propto \frac{1}{\sqrt{M}}$.The number of moles effused $n$ in time $t$ is given by $n = r 	imes t$.Therefore,the ratio of moles effused is $\frac{n_{H}}{n_{O}} = \frac{r_{H} 	imes t}{r_{O} 	imes t} = \frac{\sqrt{M_{O}}}{\sqrt{M_{H}}}$.Given $M_{H} = 2 \ g/mol$ and $M_{O} = 32 \ g/mol$.$\frac{n_{H}}{n_{O}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$.So,$n_{H} = 4 	imes n_{O}$.We are given that $n_{H} = 1/2 \ mole$ (half of the initial $1 \ mole$).$1/2 = 4 	imes n_{O} \implies n_{O} = 1/8 \ mole$.Since we started with $1 \ mole$ of oxygen,the fraction of oxygen that escapes is $1/8$.

Equal moles of hydrogen and oxygen gases are placed in a container with a pin-hole through which both can escape. What fraction of the oxygen escapes in the time required for one-half of the hydrogen to escape?

Similar Questions

For one mole of an ideal gas,increasing the temperature from $10^{\circ} C$ to $20^{\circ} C$:

The combination of plots which do not represent isothermal expansion of an ideal gas is

The radius of the $Na^{+}$ ion is $95 \ pm$ and that of $Cl^{-}$ ion is $181 \ pm$. Predict the coordination number of $Na^{+}$ in $NaCl$.

At $STP$,$0.50 \ mol$ $H_2$ gas and $1.0 \ mol$ $He$ gas:

What happens when oil is added to water? Why?

Vedclass Test Series

Exam Paper Generator

Online Exam Module