A `1.24 M` aqueous solution of `KI` has density of `1.15 g cm^(-3)`.
Answer the following questions about this solution:
The percentage composition of solute in the solution is

Options

A. `17.89`
B. `27.89`
C. `37.89`
D. `47.89`

Answer

To find the percentage composition of solute in a 1.24 M aqueous solution of KI with a density of 1.15 g/cm³, we can follow these steps:

### Step 1: Calculate the number of moles of KI in 1 liter of solution
Given that the molarity (M) of the solution is 1.24 M, this means there are 1.24 moles of KI in 1 liter of solution.

### Step 2: Calculate the mass of KI
To find the mass of KI, we need to know its molar mass. The molar mass of KI can be calculated as follows:
- Molar mass of K (Potassium) = 39 g/mol
- Molar mass of I (Iodine) = 127 g/mol
- Therefore, Molar mass of KI = 39 + 127 = 166 g/mol

Now, we can calculate the mass of KI:
\[ \text{Mass of KI} = \text{Number of moles} \times \text{Molar mass} = 1.24 \, \text{moles} \times 166 \, \text{g/gmol} = 205.84 \, \text{g} \]

### Step 3: Calculate the weight of the solution
Using the density of the solution, we can find the weight of the solution. The density is given as 1.15 g/cm³. Since we are considering 1 liter of solution, we convert liters to cubic centimeters:
\[ 1 \, \text{L} = 1000 \, \text{cm}^3 \]

Now, we can calculate the weight of the solution:
\[ \text{Weight of solution} = \text{Density} \times \text{Volume} = 1.15 \, \text{g/cm}^3 \times 1000 \, \text{cm}^3 = 1150 \, \text{g} \]

### Step 4: Calculate the percentage composition of solute in the solution
The percentage composition of the solute (KI) in the solution can be calculated using the formula:
\[ \text{Percentage composition} = \left( \frac{\text{Weight of solute}}{\text{Weight of solution}} \right) \times 100 \]
Substituting the values we calculated:
\[ \text{Percentage composition} = \left( \frac{205.84 \, \text{g}}{1150 \, \text{g}} \right) \times 100 \approx 17.89\% \]

### Final Answer
The percentage composition of solute (KI) in the solution is approximately **17.89%**.

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