if 0.00531 mol N2O effuses through an orifice in a certain period of time, how many moles of N2O3 would effuse in the same time under the same conditions?

The number moles that is able to effuse given the number of moles of another compound that effused can be determined by taking the square root of the reciprocal of the molar mass of the compounds. In this case, x/0.00531= [tex] \sqrt{ \frac{44}{76} } [/tex] where x is the number of moles of N2O3 effused. The answer is 0.00404 moles.

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Alleei Alleei · Answer 2 · 2019-10-15T11:56:16+02:00

Answer : The number of moles of [tex]N_2O_3[/tex] effused at same condition are 0.00404 moles.

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

[tex]R\propto \sqrt{\frac{1}{M}}[/tex]

or,

[tex](\frac{R_2}{R_1})=\sqrt{\frac{M_1}{M_2}}[/tex] ..........(1)

where,

[tex]R_1[/tex] = rate of effusion of [tex]N_2O[/tex] gas

[tex]R_2[/tex] = rate of effusion of [tex]N_2O_3[/tex] gas

[tex]M_1[/tex] = molar mass of [tex]N_2O[/tex] gas = 44 g/mole

[tex]M_2[/tex] = molar mass of [tex]N_2O_3[/tex] gas = 76 g/mole

Now put all the given values in the above formula 1, we get:

[tex](\frac{R_2}{R_1})=\sqrt{\frac{44g/mole}{76g/mole}}[/tex]

[tex]\frac{R_2}{R_1}=0.76[/tex]

[tex]R_2=0.76\times R_1[/tex]

Thus, the rate of effusion of [tex]N_2O_3[/tex] is 0.76 times of the rate of effusion of [tex]N_2O[/tex]

Number of moles of [tex]N_2O[/tex] effused = 0.00531 mol

Number of moles of [tex]N_2O_3[/tex] effused at same condition = [tex]0.76\times R_{N_2O}[/tex]

Number of moles of [tex]N_2O_3[/tex] effused at same condition = [tex]0.76\times 0.00531=0.00404mol[/tex]

Therefore, the number of moles of [tex]N_2O_3[/tex] effused at same condition are 0.00404 moles.