If `0.5 g` of a mixture of two metals. `A` and `B` with respective equivalent weights 12 and 9 displace `560 mL` of `H_(2)` at `STP` from an acid, the composition of the mixture is

Options

A. `40% A, 60% B`
B. `60% A, 40% B`
C. `30% A, 70% B`
D. `70% A, 30% B`

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the equivalent of H₂ displaced
Given that `560 mL` of H₂ is displaced at STP, we first need to find out how many equivalents of H₂ this volume corresponds to.

At STP, 1 mole of any gas occupies `22.4 L` or `22,400 mL`. Thus, we can find the equivalent of H₂:

\[ \text{Equivalent of H₂} = \frac{560 \text{ mL}}{11200 \text{ mL/equiv}} = \frac{560}{11200} = \frac{1}{20} \text{ equivalents} \]

### Step 2: Set up the equation for the mixture
Let the weight of metal A be \( x \) grams. Therefore, the weight of metal B will be:

\[ \text{Weight of B} = 0.5 - x \text{ grams} \]

The equivalent of A can be calculated using its equivalent weight (12):

\[ \text{Equivalent of A} = \frac{x}{12} \]

The equivalent of B can be calculated using its equivalent weight (9):

\[ \text{Equivalent of B} = \frac{0.5 - x}{9} \]

### Step 3: Write the equation for total equivalents
Since the total equivalents of the metals must equal the equivalents of H₂ displaced, we can write:

\[ \frac{x}{12} + \frac{0.5 - x}{9} = \frac{1}{20} \]

### Step 4: Solve for \( x \)
To solve this equation, we will first find a common denominator, which is `180` (LCM of 12, 9, and 20):

\[ \frac{15x}{180} + \frac{20(0.5 - x)}{180} = \frac{9}{180} \]

Now, simplifying:

\[ 15x + 10 - 20x = 9 \]

Combine like terms:

\[ -5x + 10 = 9 \]

Subtract 10 from both sides:

\[ -5x = -1 \]

Divide by -5:

\[ x = 0.2 \text{ grams} \]

### Step 5: Calculate the percentage composition
Now that we have \( x \), we can find the weight of B:

\[ \text{Weight of B} = 0.5 - 0.2 = 0.3 \text{ grams} \]

Next, we calculate the percentage of A and B in the mixture:

\[ \text{Percentage of A} = \left(\frac{0.2}{0.5}\right) \times 100 = 40\% \]

\[ \text{Percentage of B} = \left(\frac{0.3}{0.5}\right) \times 100 = 60\% \]

### Final Result
Thus, the composition of the mixture is:
- Metal A: 40%
- Metal B: 60%

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