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A cylinder of gas supplied by Bharat Petroleum is assumed to contain `14 kg` of butane. 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. `15` daysB. `20` daysC. `50` daysD. `40` days |
Answer» Correct Answer - D Molar mass of butane, `C_(4)H_(10)=12xx4+10` `=48+10=58 g mol^(-1)` `58g` of butane gives `2658 kJ` of heat energy `:. 14 kg` of butane will give heat energy `=(2658kJxx(14xx10^(3)g))/(58g)=641.5862xx10^(3)kJ` Daily energy requirement for cooking `=20,000 kJ` `=2xx10^(4) kJ "day"^(-1)` `:. ` No. of days cylinder will last `=(641.5862xx10^(3)kJ)/(2xx10^(4)kJ"day^(-1))=32.08"days"` |
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