At T K,equal weights of H_2, D_2,and T_2 are | vedclass

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

Question

(B) Let the equal weight of each gas be $w \ g$. The molar masses are $M(H_2) = 2 \ g/mol$,$M(D_2) = 4 \ g/mol$,and $M(T_2) = 6 \ g/mol$.Number of mol

Vedclass · Accepted Answer

$0.18 : 0.27 : 0.54$

Vedclass · Answer

(B) Let the equal weight of each gas be $w \ g$. The molar masses are $M(H_2) = 2 \ g/mol$,$M(D_2) = 4 \ g/mol$,and $M(T_2) = 6 \ g/mol$.Number of moles: $n(H_2) = \frac{w}{2}$,$n(D_2) = \frac{w}{4}$,$n(T_2) = \frac{w}{6}$.Total moles $n_{	ext{total}} = w(\frac{1}{2} + \frac{1}{4} + \frac{1}{6}) = w(\frac{6+3+2}{12}) = \frac{11w}{12}$.Mole fractions: $x(H_2) = \frac{w/2}{11w/12} = \frac{6}{11} \approx 0.545$,$x(D_2) = \frac{w/4}{11w/12} = \frac{3}{11} \approx 0.273$,$x(T_2) = \frac{w/6}{11w/12} = \frac{2}{11} \approx 0.182$.Since partial pressure $P_i = x_i P_{	ext{total}}$,the ratio of partial pressures $P(T_2) : P(D_2) : P(H_2)$ is equal to the ratio of their mole fractions $x(T_2) : x(D_2) : x(H_2) = 0.18 : 0.27 : 0.54$.

At $T \ K$,equal weights of $H_2, D_2$,and $T_2$ are present in a closed vessel. The pressure exerted by this gaseous mixture is $P \ atm$. The ratio of partial pressures of $T_2, D_2$,and $H_2$ is approximately. ($H, D$ and $T$ are isotopes of hydrogen).

Similar Questions

$A$ $10 \ L$ vessel contains $1 \ mole$ of an ideal gas with pressure of $P \ atm$ and temperature of $T \ K$. The vessel is divided into two equal parts. The pressure (in $atm$) and temperature (in $K$) in each part is respectively:

Dalton's law of partial pressure is not applicable for which of the following mixtures?

The total pressure exerted by a mixture of non-reactive gases is equal to the sum of the partial pressures of individual gases. This is known as which law?

$12 \ cm^3$ of $SO_{2(g)}$ diffused through a porous membrane in $1 \ minute$. Under similar conditions $120 \ cm^3$ of another gas diffused in $5 \ minutes$. The molar mass of the gas in $g \ mol^{-1}$ is

$50 \ mL$ of hydrogen diffuses out through a small hole from a vessel in $20 \ min$. The time needed for $40 \ mL$ of oxygen to diffuse out is ............. $min$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module