Two oxides of a certain metal were seperately heated in hydrogen till water is produced. It was observed that 1 gm of each oxide gave 0.125 gm and 0.2263 grams of water respectively. This law states that it illustrate law of

Options

A. Definite proportions
B. Multiple proportions
C. Reciprocal proportions
D. Conservation of mass

Explanation

To solve the problem, we need to analyze the data provided and apply the concept of the law of multiple proportions. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Problem
We have two oxides of a metal, and when heated in hydrogen, they produce water. The weights of water produced from each oxide are given as 0.125 g and 0.2263 g respectively. We need to determine what law this illustrates.

### Step 2: Calculate the Mass of Oxygen in Water
The molecular weight of water (H₂O) is 18 g/mol (2 g from hydrogen and 16 g from oxygen). We can find the mass of oxygen in the water produced from each oxide using the following formula:

\[ \text{Mass of Oxygen} = \left(\frac{16 \text{ g}}{18 \text{ g}}\right) \times \text{Mass of Water} \]

#### For the first oxide:
\[ \text{Mass of Oxygen}_1 = \left(\frac{16}{18}\right) \times 0.125 \text{ g} = 0.1111 \text{ g} \]

#### For the second oxide:
\[ \text{Mass of Oxygen}_2 = \left(\frac{16}{18}\right) \times 0.2263 \text{ g} = 0.2011 \text{ g} \]

### Step 3: Calculate the Mass of the Metal
Since the mass of each oxide is given as 1 g, we can find the mass of the metal in each case by subtracting the mass of oxygen from the total mass of the oxide.

#### For the first oxide:
\[ \text{Mass of Metal}_1 = 1 \text{ g} - 0.1111 \text{ g} = 0.8889 \text{ g} \]

#### For the second oxide:
\[ \text{Mass of Metal}_2 = 1 \text{ g} - 0.2011 \text{ g} = 0.7989 \text{ g} \]

### Step 4: Calculate the Ratios of Oxygen to Metal
Now, we find the ratio of the mass of oxygen that combines with 1 g of metal for both cases.

#### For the first oxide:
\[ \text{Ratio}_1 = \frac{0.1111 \text{ g}}{0.8889 \text{ g}} \approx 0.1249 \]

#### For the second oxide:
\[ \text{Ratio}_2 = \frac{0.2011 \text{ g}}{0.7989 \text{ g}} \approx 0.2517 \]

### Step 5: Find the Ratio of the Masses of Oxygen
Now, we can find the ratio of the two masses of oxygen:

\[ \text{Ratio of Oxygen} = \frac{0.1249}{0.2517} \approx 0.496 \]

To express this in whole numbers, we can multiply both sides by 2 to get approximately 1:2.

### Step 6: Conclusion
The ratio of the masses of oxygen that combine with a fixed mass of metal (1 g) is in the ratio of 1:2. This illustrates the **Law of Multiple Proportions**, which states that when two elements form more than one compound, the ratios of the masses of the second element that combine with a fixed mass of the first element can be expressed as small whole numbers.

### Final Answer
The law illustrated by the given data is the **Law of Multiple Proportions**.

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