1.

A and B are two hydrocarbons. A and B are heated separated in excess of oxygen when 0.028 g of A gave 44.8mL CO_(2) and 0.44g of B gave 67.2mL CO_(2) at NTP. Show that the results are in agreement with law of multiple proportions.

Answer»

Solution :Determine the masses of `CO_(2)` at NTP and then masses of carbon.
(A) Mass of `CO_(2)=(44)/(22400)xx44.8=0.088g`.
(B) Mass of `CO_(2)=(44)/(22400)xx67.2=0.132g`
mass of carbon=0.036g, mass of hydrogen =0.008 g Thus, the masses of carbon combining with same mass of hydrogen are in the ratio of 4:3 which is a SIMPLE ratio. Hence, law of multiple proportions is followed.


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