mixture of KBr and NaBr weighing 0.563 g was treated with aqueous `AgNO_(3)` and all the bromide ion was recovered as 0.975 g of pure AgBr. What fraction of total mass is NaBr in the sample ?

Options

A. 0.61
B. 0.22
C. 0.83
D. 0.45

Explanation

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

### Step 1: Understand the Reaction
When KBr and NaBr are treated with AgNO3, they react to form AgBr and the respective nitrates (KNO3 and NaNO3). The reaction can be represented as:
- KBr + AgNO3 → AgBr + KNO3
- NaBr + AgNO3 → AgBr + NaNO3

### Step 2: Define Variables
Let:
- \( x \) = mass of KBr in the mixture (in grams)
- \( 0.563 - x \) = mass of NaBr in the mixture (in grams)

### Step 3: Calculate Molar Masses
- Molar mass of KBr = 39 (K) + 80 (Br) = 119 g/mol
- Molar mass of NaBr = 23 (Na) + 80 (Br) = 103 g/mol
- Molar mass of AgBr = 108 (Ag) + 80 (Br) = 188 g/mol

### Step 4: Set Up the Equation
The total mass of AgBr produced from both KBr and NaBr can be expressed in terms of the masses of KBr and NaBr:
\[ \frac{188}{119} \cdot x + \frac{188}{103} \cdot (0.563 - x) = 0.975 \]

### Step 5: Substitute Known Values
Substituting the known values into the equation:
\[ \frac{188}{119} \cdot x + \frac{188}{103} \cdot (0.563 - x) = 0.975 \]

### Step 6: Solve for \( x \)
1. Calculate the coefficients:
- \( \frac{188}{119} \approx 1.577 \)
- \( \frac{188}{103} \approx 1.826 \)

2. Rewrite the equation:
\[ 1.577x + 1.826(0.563 - x) = 0.975 \]

3. Expand and simplify:
\[ 1.577x + 1.026 - 1.826x = 0.975 \]
\[ -0.249x + 1.026 = 0.975 \]
\[ -0.249x = 0.975 - 1.026 \]
\[ -0.249x = -0.051 \]
\[ x = \frac{-0.051}{-0.249} \approx 0.205 \text{ g (for KBr)} \]

### Step 7: Calculate Mass of NaBr
Now, calculate the mass of NaBr:
\[ \text{Mass of NaBr} = 0.563 - x = 0.563 - 0.205 \approx 0.358 \text{ g} \]

### Step 8: Calculate the Fraction of NaBr
The fraction of NaBr in the mixture is given by:
\[ \text{Fraction of NaBr} = \frac{\text{Mass of NaBr}}{\text{Total mass}} = \frac{0.358}{0.563} \approx 0.636 \]

### Step 9: Final Answer
Thus, the fraction of total mass that is NaBr in the sample is approximately:
\[ \text{Fraction of NaBr} \approx 0.636 \]