Apne doubts clear karein ab Whatsapp par bhi. Try it now.

CLICK HERE

Loading DoubtNut Solution for you

Watch 1000+ concepts & tricky questions explained!

7.0 K+ views | 300+ people like this

Like ShareShare

Answer Text

Solution :

Suppose there are `n_(1)` moles of hydrogen and `n_(2)` moles of oxygen in the mixture. The pressyre of the mixture will be <br> `p=(n_(1)RT)/(V)+n_(2)RT/(V)=(n_(1)+n_(2))(RT)/(V)` <br> or, `100xx10^(3)Pa=(n_(1)+n_(2))(8.3JK^(-1)mol^(-1)(300K))/(2000xx10^(-6)m^(-3)) <br> or, n_(1)+n_(2)=0.08mol`. ...(i) <br> The mass of the mixture is <br> `n_(1)xx2g mol^(-1)+n_(2)xx32g mol^(-1)=0.76g` <br> or, `n_(1)+16n_(2)=0.38mol`. ...(ii) <br> from (i) and(ii), <br> `n_(1)=0.06 mol and n_(2)=0.02 mol`. <br> the mass of hydrogen = `0.06xx2g =0.12g` and the mass of oxygen `=0.02xx32g=0.64g`.