What volume of water (in mL) should be added to 50 ml of `HNO_(3)` having density `1.5 g` `ml^(-)` and `63.0%` by weight to have one molar solution

Answer

To solve the problem of determining what volume of water should be added to 50 mL of HNO₃ to create a one molar solution, we will follow these steps:

### Step 1: Calculate the mass of HNO₃ in the solution
Given:
- Volume of HNO₃ = 50 mL
- Density of HNO₃ = 1.5 g/mL

Using the formula for mass:
\[ \text{Mass} = \text{Density} \times \text{Volume} \]

Substituting the values:
\[ \text{Mass} = 1.5 \, \text{g/mL} \times 50 \, \text{mL} = 75 \, \text{g} \]

### Step 2: Calculate the mass of pure HNO₃ in the solution
Given:
- The solution is 63% HNO₃ by weight.

To find the mass of pure HNO₃:
\[ \text{Mass of pure HNO}_3 = \text{Total mass} \times \left( \frac{\text{Percentage}}{100} \right) \]
\[ \text{Mass of pure HNO}_3 = 75 \, \text{g} \times \left( \frac{63}{100} \right) = 47.25 \, \text{g} \]

### Step 3: Calculate the number of moles of HNO₃
The molar mass of HNO₃ is approximately 63 g/mol.

Using the formula for moles:
\[ \text{Moles} = \frac{\text{Mass}}{\text{Molar mass}} \]
\[ \text{Moles of HNO}_3 = \frac{47.25 \, \text{g}}{63 \, \text{g/mol}} \approx 0.75 \, \text{mol} \]

### Step 4: Determine the final volume needed for a 1 M solution
For a 1 M solution, the number of moles of solute (HNO₃) should equal the volume in liters.

Since we have 0.75 moles, the volume in liters needed is:
\[ \text{Volume} = \text{Moles} \div \text{Concentration} = 0.75 \, \text{mol} \div 1 \, \text{mol/L} = 0.75 \, \text{L} = 750 \, \text{mL} \]

### Step 5: Calculate the volume of water to be added
We initially have 50 mL of HNO₃, so the volume of water to be added is:
\[ \text{Volume of water} = \text{Final volume} - \text{Initial volume} \]
\[ \text{Volume of water} = 750 \, \text{mL} - 50 \, \text{mL} = 700 \, \text{mL} \]

### Final Answer
The volume of water that should be added is **700 mL**.

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