Equal weight of CO and CH_4 are mixed together in an empty container at 300 ~K . The fraction of total pressure exerted by CH_4 is

Correct answer: B. (7)/(11) Let the weight of CO= weight of CH_4=a ~g Moles of CO=(a)/(28) Moles of CH_4=(a)/(16) Total moles =(a)/(28)+(a)/(16)x_ CH_4 = ((a)/(16))/((a)/(28)+(a)/(16))=(a)/(16) × (28 × 16)/(44 a)=(14)/(22)=(7)/(11) Fraction of pressure exerted by CH_4=7 / 11 .

AIIMS 2010

Exam: AIIMS
Year: 2010
Subject: Chemistry
Chapter: States of Matter

AIIMS 2010 · Chemistry · States of Matter

Equal weight of CO and CH_4 are mixed together in an empty...

Equal weight of

\mathrm{CO}

and

\mathrm{CH}_4

are mixed together in an empty container at

300 \mathrm{~K}

. The fraction of total pressure exerted by

\mathrm{CH}_4

Options

A $\frac{16}{17}$
B $\frac{7}{11}$
C $\frac{8}{9}$
D $\frac{5}{16}$

Correct answer

B. $\frac{7}{11}$

Step-by-step solution

Let the weight of

\mathrm{CO}=

weight of

\mathrm{CH}_4=a \mathrm{~g}

Moles of

\mathrm{CO}=\frac{a}{28}

Moles of

\mathrm{CH}_4=\frac{a}{16}

Total moles

=\frac{a}{28}+\frac{a}{16}x_{\mathrm{CH}_4

\frac{\frac{a}{16}}{\frac{a}{28}+\frac{a}{16}}=\frac{a}{16} \times \frac{28 \times 16}{44 a}=\frac{14}{22}=\frac{7}{11}\therefore

Fraction of pressure exerted by

\mathrm{CH}_4=7 / 11

Practice States of Matter on Quantrex Academy →

More from States of Matter