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 weight. State the law illustrated by these chemical combinations.

• A. Law of multiple proportion
• B. Law of constant composition
• C. Law of reciprocal proportion
• D. None of the above

Answer 1

Answer: (B)

Law of constant composition

First case:
Weight of $10$ mL of $H_2$ at STP $=\frac{10\times 2}{22400}=0.0008$ g

Weight of $5$ mL of $O_2$ at STP $=\frac{5\times 32}{22400}=0.0007$ g

Second case:
Weight of $200$ mL of $H_2$ at STP $=\frac{200\times 2}{22400}=0.0178$ g
Weight of oxygen taken away from cupric oxide by $200$ mL or $0.0178$ 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 $=\frac{0.144\times 0.0008}{0.0178}=0.0065\approx 0.007$ 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.

Hence, it illustrates the law of constant composition.