InterviewSolution
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InterviewSolution
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A cylinder of gas supplied by Bharat Petroleum is assumed to contain `14 kg` of butance. If a normal family requires 20,000 kJ of energy per day for cooking, butane gas in the cylinder lasts `(Delta_(C ) H^(@)` of `C_(4) H_(10) = - 2658 kJ mol^(-1))`A. 20B. 50C. 40D. 32 |
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Answer» Correct Answer - D Number of days cylinder will last `("Energy available from 14 kg butane")/("Energy requirement per day")` Energy released out by the combustion of 14 kg of butane is = (Number of moles of butane) `((14 xx 10^(3)g)/(58 g mol^(-1))) (2658 kJ mol^(-1))` `= 641.6 xx 10^(3) kJ` Note that `n_(C_(4)H_(10)) = Mass_(C_(4)H_(10))//"Molar mass"_(C_(4)H_(10))` and molar mass of `C_(4) H_(10` is `58 g mol^(-1)`. Thus, the number of days cylinder will last `= (641.6 xx 10^(3) kJ)/(20 xx 10^(3) kJ day^(-1))` `=32.08 days` |
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