A bottle of `H_(3)PO_(4)` solution contains 70% acid. If the density of the solution is `1.54 g cm^(-3)` , the volume of the `H_(3)PO_(4)` solution required to prepare 1L of IN solution is .

Options

A. 90 mL
B. 45mL
C. 30mL
D. 23mL

Answer

To solve the problem of determining the volume of a 70% `H₃PO₄` solution required to prepare 1L of a 1N solution, we can follow these steps:

### Step 1: Calculate the Molarity of the `H₃PO₄` Solution
We use the formula for molarity when density and mass percent are involved:

\[ \text{Molarity (M)} = \frac{10 \times \text{Density (g/cm}^3\text{)} \times \text{Mass percent}}{\text{Molecular weight (g/mol)}} \]

Given:
- Density = 1.54 g/cm³
- Mass percent = 70%
- Molecular weight of `H₃PO₄` = 98 g/mol

Substituting the values:

\[ \text{Molarity} = \frac{10 \times 1.54 \times 70}{98} \]

Calculating this gives:

\[ \text{Molarity} = \frac{10 \times 1.54 \times 70}{98} = \frac{1078}{98} \approx 11 \text{ M} \]

### Step 2: Determine the Normality of the `H₃PO₄` Solution
The normality (N) is related to molarity (M) by the equation:

\[ \text{Normality (N)} = n \times \text{Molarity (M)} \]

Where `n` is the number of equivalents (or the n-factor). For `H₃PO₄`, which is a triprotic acid, `n` = 3.

Thus,

\[ \text{Normality} = 3 \times 11 = 33 \text{ N} \]

### Step 3: Use the Normality to Find the Volume Required
We can use the formula:

\[ n_1 \times V_1 = n_2 \times V_2 \]

Where:
- \( n_1 \) = Normality of the `H₃PO₄` solution = 33 N
- \( V_1 \) = Volume of the `H₃PO₄` solution required (what we need to find)
- \( n_2 \) = Normality of the desired solution = 1 N
- \( V_2 \) = Volume of the desired solution = 1 L = 1000 mL

Substituting the values into the equation:

\[ 33 \times V_1 = 1 \times 1000 \]

Solving for \( V_1 \):

\[ V_1 = \frac{1000}{33} \approx 30.3 \text{ mL} \]

### Conclusion
The volume of the `H₃PO₄` solution required to prepare 1L of a 1N solution is approximately **30 mL**.