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7. 1.375 g of cupric oxide on reduction in hydrogen gas gives 1.098 g of copper. In another experiment, 1.179 g of metallic copper produced 1.476 g of copper oxide. Show that these results illustrate the law of constant proportions.

Solution

Mass of copper oxide taken in experiment 1. = 1.375 g

Mass of copper obtained = 1.098 g

Mass of copper oxide produced in experiment 2 = 1.476 g

Mass of copper used = 1.179 g

Since the percentage of copper in the two samples of copper oxide is the same, the law of definite proportion is verified.

8. Carbon and oxygen are known to form two compounds. The carbon content in one of these is 42.9% while in the other it is 27.3%. Show that this data is in agreement with the law of multiple proportions.

Solution

For the first compound, mass % of C = 42.9

Mass % of O = 57.1

Thus, 42.9 g of C reacts with 57.1 g of oxygen

=1.33 g of oxygen

For the second compound, mass % of C = 27.3

Mass % of O = 72.7

Thus, 27.3 g of C reacts with 72.7 g of oxygen

= 2.66 g of oxygen

The ratio of oxygen masses, which combine with 1 g of C is 1.33:2.66 or 1:2

This is a simple ratio and supports law of multiple proportions.9. Hydrogen sulphide contains 94.11% sulphur. Sulphur dioxide contains 50% oxygen. Water contains 11.11% hydrogen. Show that the results are in agreement with the law of reciprocal proportions.

Solution

In H₂S, 100-94.11 = 5.89 g of hydrogen combines with 94.11 g of sulphur. So, 1g of hydrogen combines with

of sulphur.

In H₂O, 100-11.11 = 88.89 g of oxygen combines with 11.11 g of hydrogen. So, 1g of hydrogen combines with

of oxygen.

Ratio of the masses of sulphur (H₂S) and oxygen (H₂O)

Ratio of the masses of sulphur and oxygen (SO₂)

The first ratio is double of the second, which is a simple ratio. This illustrates the law of reciprocal proportions.