Question

$10 mL$ of hydrogen combines with $5 mL$ of oxygen to yield water. When $200 mL$ of hydrogen at $STP$ is passed over heated $CuO$, the $CuO$ loses $0.144 g$ of its wieght. Do these result correspond to the law of constant composition?

Answer 1

Correct Answer - A
First method
First case
Weight of $10 mL$ of $H(2)$ at $STP = (10 xx 2)/(22400) = 0.0008 g$
Weight of $5 mL$ of $O(2)$ at $STP = (5 xx 32)/(22400) = 0.0007 g$
Second case
Weight of $200 m$L of $H(2)$ at $STP = (200 xx 2)/(22400) = 0.1178 g$
Weight of oxygen taken away from copic oxide by $200 mL$ of $0.0178 g$ of $H(2)$ at $STP = 0.144 g$
Weight of $O(2)$ that combines with $0.0008 g$ of $H(2)$ in first case $= (0.144 xx 0.0008)/(0.0178) = 0.0065 ~~ 0.007 g$
Thus, the weights of $O(2)$ that combine with same weight of $H(2)$ is the same in two cases. Hence, the law of constant composition is proved.
Second method
First case
$underset({:(1 mL),(10 mL),(200 mL):})(H(2)) + underset({:((1)/(2) mL),(5 mL),(100 mL):})((1)/2) O(2) rarr H(2) O$
Second case
Weight of oxygen taken away from cupric oxide by $200 mL$ of $H(2)$ at $STP = 0.144 g$
$:.$ volume of $0.144 g of O(2) = (22400 xx 0.144)/(32)$
$= 100.8 mL$
So the volume of $O(2)$ that combines with $H_(2)$ in both the cases is same. Hence, the law of constant composition is verifed.