A metal forms two oxides. The higher oxide contains 80% metal. 0.72 g of the lower oxide gave 0.8 g of higher oxide when oxidiesd. Then the ratio of the weight of oxygen that combines with the fixed weight of the metal in the two oxides will be

Options

A. `2 : 3`
B. `1 : 2`
C. `4 : 5`
D. `3 : 2`

Answer

To solve the problem step by step, we will analyze the information given and perform the necessary calculations.

### Step 1: Determine the mass of oxygen used in the oxidation process.
- We know that 0.72 g of the lower oxide produces 0.8 g of the higher oxide upon oxidation.
- The mass of oxygen used can be calculated as follows:
\[ \text{Mass of oxygen used} = \text{Mass of higher oxide} - \text{Mass of lower oxide} = 0.8 \, \text{g} - 0.72 \, \text{g} = 0.08 \, \text{g} \]

### Step 2: Calculate the mass of metal in the higher oxide.
- The higher oxide contains 80% metal. Therefore, the mass of metal in the higher oxide can be calculated as:
\[ \text{Mass of metal in higher oxide} = 80\% \text{ of } 0.8 \, \text{g} = \frac{80}{100} \times 0.8 \, \text{g} = 0.64 \, \text{g} \]

### Step 3: Determine the mass of oxygen in the higher oxide.
- Since the higher oxide is composed of metal and oxygen, we can find the mass of oxygen in the higher oxide:
\[ \text{Mass of oxygen in higher oxide} = \text{Mass of higher oxide} - \text{Mass of metal in higher oxide} = 0.8 \, \text{g} - 0.64 \, \text{g} = 0.16 \, \text{g} \]

### Step 4: Calculate the ratio of the weight of oxygen that combines with the fixed weight of metal in the two oxides.
- We have found that the mass of oxygen used with the lower oxide is 0.08 g and the mass of oxygen in the higher oxide is 0.16 g.
- To find the ratio of the weight of oxygen that combines with the fixed weight of metal in the two oxides, we will take the ratio of the two masses:
\[ \text{Ratio of oxygen} = \frac{\text{Oxygen with lower oxide}}{\text{Oxygen with higher oxide}} = \frac{0.08 \, \text{g}}{0.16 \, \text{g}} = \frac{1}{2} \]

### Conclusion
The ratio of the weight of oxygen that combines with the fixed weight of metal in the two oxides is **1:2**.

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