Question

Two elements A and B combine to form compound X and Y. For the fixed mass of A, masses of B combined for the compounds X and Y are in 3 : 7 ratio. If in compound X, 4 g of A combines with 12 g B, then in compound Y, 8 g of A will combine with...................g of B.

• A. 24
• B. 56
• C. 28
• D. 8

Answer 1

Answer: (B)

56

Answer 2

Let the two compounds formed by elements A and B be X and Y.

In compound X, 4 g of A combines with 12 g of B. The ratio of mass of B to mass of A in compound X is $\frac{\text{Mass of B}}{\text{Mass of A}} = \frac{12 \text{ g}}{4 \text{ g}} = 3$.

The problem states that for a fixed mass of A, the masses of B combined for compounds X and Y are in the ratio 3 : 7. Let's consider a fixed mass of A, say 4 g (as given for compound X). For this fixed mass of A (4 g), the mass of B in compound X is 12 g. Let the mass of B that combines with 4 g of A in compound Y be $m_Y$. According to the given ratio:

$\frac{\text{Mass of B in X (for 4g A)}}{\text{Mass of B in Y (for 4g A)}} = \frac{3}{7}$

$\frac{12 \text{ g}}{m_Y} = \frac{3}{7}$

$m_Y = \frac{12 \text{ g} \times 7}{3} = 4 \text{ g} \times 7 = 28 \text{ g}$. So, in compound Y, 4 g of A combines with 28 g of B.

Now, we need to find the mass of B that combines with 8 g of A in compound Y. According to the Law of Definite Proportions, the ratio of masses of elements in a chemical compound is fixed. In compound Y, the ratio of mass of B to mass of A is constant:

$\frac{\text{Mass of B}}{\text{Mass of A}} = \frac{28 \text{ g}}{4 \text{ g}} = 7$.

If 8 g of A combines with $m'_B$ g of B in compound Y, then:

$\frac{m'_B}{8 \text{ g}} = 7$

$m'_B = 7 \times 8 \text{ g} = 56 \text{ g}$.

Therefore, in compound Y, 8 g of A will combine with 56 g of B.