The ratio of O2/H2 in the balloon the next day is 1.4 .
Given ,
For a combustion decomposition , a professor's assistant fills a balloon with equal molar amounts of H2 and O2 , but the demonstration has to be postponed until the next day .
During night , both the gases leak through pores in the balloon .
The % of H2 leaks = 35%
Moles of H2 leaks = 0.35 mol
moles of H2 remaining = 0.65 mol
The molecular mass of O2 = 32g
The molecular mass of H2 =2g
Thus, the rate of diffusion of O2 = R
and the rate of diffusion of H2 = r
Now , Graham's law of diffusion ,
[tex]R/r = \sqrt{\frac{M( hydrogen ) }{M( oxygen)} } =\sqrt{\frac{2}{32} } =\frac{1}{4}[/tex]
thus the diffusion rate of O2/H2 is 1/4 .
From this we get to know that the % of O2 leaks is 1/4 of H2 leaks .
% of O2 leaks = 1/4 *35% = 8.8%
Moles of O2 leaks = 0.088 mol
Thus , the remaining O2 in the balloon = 0.912 mol
Then , the ratio of O2/H2 in the balloon the next day = 0.912/0.65 =1.4
Hence , the ratio of O2/H2 the next day is 1.4 .
Learn more about Graham's law of diffusion here :
brainly.com/question/12415336
#SPJ4
