A sample of ethane `(C_(2)H_(6))` gas has the same mass a `1.5 xx 10^(20)` moleucles of methane `(CH_(4))`. How many `C_(2)H_(6)`. How many `C_(2)H_(6)` molecules does the sample of gas contain?

Explanation

No. of molecules of methane `(CH_(4))=1.5xx10^(20)`
`"Mass of one moelcule of methane"=("Gram molecular mass")/("Avogadro's number")=(16g)/(N_(A))`
`"Mass of one molecule of methane"=("Gram molecular mass")/("Avogadro's number")=(16g)/(N_(A))`
`"Mass of ethane"(C_(2)H_(6))"molecules"=3.985xx10^(-3)g`
`"Gram molar mass of ethane"=2xx12+6xx1=30g`
`"No. of molecules of ethane present"=(N_(A))/((30g))xx(3.985xx10^(-3)g)`
`=((6.022xx10)^(23)xx(3.985xx10^(-3)g))/((30g))=0.8xx10^(20)"molecules"`