The molality of `1 L` solution with `x% H_(2)SO_(4)` is equal to 9. The weight of the solvent present in the solution is `910 g`. The value of `x` in g per 100 mL is

Options

A. 90
B. 80.3
C. 49.2
D. 9

Explanation

To solve the problem step by step, we will use the information provided in the question and the definition of molality.

### Step 1: Understand the definition of molality
Molality (m) is defined as the number of moles of solute per kilogram of solvent. The formula for molality is:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]

### Step 2: Given data
- Molality (m) = 9
- Volume of solution = 1 L
- Weight of solvent = 910 g = 0.910 kg (since 1 kg = 1000 g)
- Molecular weight of H₂SO₄ = 98 g/mol

### Step 3: Calculate the weight of solute using the molality formula
Using the formula for molality:
\[ 9 = \frac{\text{weight of solute (W)}}{98 \times 0.910} \]

Rearranging the equation to find W (weight of solute):
\[ W = 9 \times 98 \times 0.910 \]

### Step 4: Calculate W
Now, we can calculate W:
\[ W = 9 \times 98 \times 0.910 \]
\[ W = 9 \times 89.18 \]
\[ W = 802.62 \text{ g} \]

### Step 5: Convert weight of solute to grams per 100 mL
Since we need the value of x in grams per 100 mL, we need to convert the weight of solute from grams per liter to grams per 100 mL:
\[ \text{Weight of solute per 100 mL} = \frac{802.62 \text{ g}}{10} = 80.262 \text{ g} \]

### Step 6: Round off the answer
Rounding off to three significant figures:
\[ x \approx 80.26 \text{ g per 100 mL} \]

### Final Answer
The value of x in grams per 100 mL is approximately **80.3 g per 100 mL**.

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