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CBSE Class 11-science Answered

13th?
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Asked by smanishkumar2002 | 04 Aug, 2018, 05:41: AM
answered-by-expert Expert Answer
Total space amount space of space gases space in space the space mixture space comma

equals PV over RT

equals fraction numerator 6 cross times 3 over denominator 0.0821 cross times 300 end fraction

equals space 0.7308 space mol

therefore The space amount space of space unknown space gas space in space the space mixture equals 0.7308 space minus space 0.7

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 0.0308 space mol

Now space

straight r subscript 1 over straight r subscript 2 space equals fraction numerator Amount space oof space hydrogen over denominator Aount space of space unknown space gas end fraction

space space space space space space equals fraction numerator 0.7 over denominator 0.0308 end fraction

we space know space that comma

straight r subscript 1 over straight r subscript 2 equals square root of straight M subscript 2 over straight M subscript 1 end root

open parentheses straight r subscript 1 over straight r subscript 2 close parentheses squared space equals straight M subscript 2 over straight M subscript 1

straight M subscript 2 equals end subscript open parentheses straight r subscript 1 over straight r subscript 2 close parentheses squared cross times straight M subscript 1

space space space space space space equals open parentheses fraction numerator 0.7 over denominator 0.0308 end fraction close parentheses squared cross times open parentheses 2 close parentheses

space space space space space space equals 1033 space straight g divided by mol
Answered by Varsha | 05 Aug, 2018, 08:51: PM
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