1nope......what's your logic ?
A container has oxygen and hydrogen. Some weight of gases escape. Same weight of methane as that of escaped gases is added to the container so that the total pressure of the mixture remains same. Then what percent of the escaped gases by weight of hydrogen escaped ?
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7 Answers
11is the answer 6.67 %.....???
first confirm then i'll post the solution...
29Let the number of moles of Hydrogen gas escaped be n1 and the number of moles of Oxygen gas escaped be n2 ..
since the pressure remains the same so n1 + n2 = y
and acc to the other condition given we have
2n1 + 32n2 = 16y ..
solving these equations u will get
n1 = 16y/30 and n2 = 14y/30
now we have to find
\frac{\frac{16y*2}{30}}{\frac{16y*2}{30}+\frac{14y*32}{30}} * 100
so from here u will get 6.67%
11See ...
let no & nH be the initial moles of oxygen and hydrogen in the container...
let n'o & n'H be the amount of moles escaping... of oxygen & hydrogen respectively...
let volume be V
so Pi = (no + nH)RTV
now ATQ,
nCH4 = (32 n'o + 2 n'H)16
nCH4 = 2 n'o + n'H8
so final moles after addition of methane..
= (no - n'o) + (nH - n'H) +(2 n'o + n'H8)
= no + nH + n'o - 7n'H8
so final pressure..
Pf = (no + nH + n'o - 7n'H8)R TV
so Pi = Pf
(no + nH)RTV = (no + nH + n'o - 7n'H8)R TV
solving we get...
n'o = 7n'H8
now m = M n (m=weight , M = molar mass , n= moles)
so
momH = Mo noMH nH
momH = 32 x 72 x 8 = 14
or mHmo = 1 14
now..
mHmH + mo = 1 1+ 14 = 115
or
mHmH + mo x 100 = 115 x 100
or %hydrogen by weight in the gas escaping=100/15= 6.67%
answer = 6.67 %... [1]
1THANKS A LOT GUYS.......IT CAME IN OUR EXAM TODAY.....WHAT LUCK !!!!!!