US UKCIWKOI. I u VGTV 9. 10 ml of a...
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US UKCIWKOI. I u VGTV 9. 10 ml of a gaseous hydrocarbon was burnt completely in 80 mL of O2 at NTP. The remaining gas occupied 70 mL at NTP. This volume became 50 mL on treatment with KOH solution, what is the formula of the hydrocarbon?

JEE/Engineering Exams
Chemistry
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1 ( c_{2} times H_{4}+0_{2} longrightarrow c_{0}_{2}+n_{2} 0 ) volume of ( ((0,2+cup text { nreackol } 0,2) text { : } 70 ) m Coz is cissorbed by kou so this 70 me mix is treated it becems 50 millit means volume up ( mathrm{CO}_{2} ). Was 20 m1. votume of 0 , reacts dra reaction ( 20-50 ) ( 3^{circ} mathrm{m} / ) wotume of hydrocerbon ( =10 mathrm{m} ) APP'ying PoAC on Carbunarony ( x^{2} times x cdot 0.9 ) moles ( 2 cos n y ) ( cdot(operatorname{nos}) ) [ x times 10=20 ] ( mid cos theta(0) ) ( x=2 ) ( rightarrow ) APPIY POAC OST ATAOMR. ( 4 cdot n_{0} cdot v_{4} ) moles ( partial gamma_{c} times h_{7}=2 times n_{0} cdot v_{j} ) moles ( q ) ( rightarrow ) APPTY POAC ON O ATOMS ( m_{2} 0 ) ( cos ) ( 2 times x_{0} cdot 15 ) moles of ( 0.2=frac{y}{2} times 200.08 ) ( theta_{5}
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