Specific gravity of a salt solution is 1.025. Quantity of the solution required that consist of 1 mole of solute `(M^(@) = 60)` if molarity of the original solution is 0.9M.

Options

A. 1008 g
B. 1138 g
C. 1492 g
D. 997 g

Explanation

To solve the problem step by step, we will follow the given information and apply the relevant formulas.

### Step 1: Understand the given data
- Specific gravity of the salt solution = 1.025
- Molarity of the original solution = 0.9 M
- Molecular weight of the solute (M) = 60 g/mol
- We need to find the quantity of the solution that contains 1 mole of solute.

### Step 2: Calculate the density of the solution
Specific gravity is defined as the ratio of the density of a substance to the density of water. Since the density of water is approximately 1 g/mL, the density of the solution can be calculated as:
\[ \text{Density of solution} = \text{Specific Gravity} \times \text{Density of water} = 1.025 \, \text{g/mL} \]

### Step 3: Calculate the volume of the solution required for 1 mole of solute
Using the formula for molarity (M):
\[ \text{Molarity (M)} = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}} \]
Given that we need 1 mole of solute and the molarity is 0.9 M, we can rearrange the formula to find the volume:
\[ \text{Volume} = \frac{\text{Number of moles}}{\text{Molarity}} = \frac{1 \, \text{mol}}{0.9 \, \text{M}} = \frac{10}{9} \, \text{L} \]

### Step 4: Convert volume from liters to milliliters
To convert the volume from liters to milliliters:
\[ \text{Volume in mL} = \frac{10}{9} \, \text{L} \times 1000 \, \text{mL/L} = \frac{10000}{9} \, \text{mL} \approx 1111.11 \, \text{mL} \]

### Step 5: Calculate the mass of the solution using density
Using the density formula:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]
We can rearrange this to find the mass:
\[ \text{Mass} = \text{Density} \times \text{Volume} \]
Substituting the values:
\[ \text{Mass} = 1.025 \, \text{g/mL} \times \frac{10000}{9} \, \text{mL} = 1.025 \times \frac{10000}{9} \approx 1138 \, \text{g} \]

### Step 6: Conclusion
The quantity of the solution required that consists of 1 mole of solute is approximately **1138 g**.