105 mL of pure water at `4^oC` is saturated with `NH_3` gas yielding a solution of density `0.9gmL^(−1)`and containing 30% `NH_3` by mass. The volume (in litres) of `NH_3` gas at `4^oC` and 775 mm of Hg.

Answer

To solve the problem step by step, we will follow the outlined approach in the video transcript.

### Step 1: Determine the mass of NH₃ in the solution
Given that the solution is 30% NH₃ by mass, we can calculate the mass of NH₃ in a certain mass of the solution.

For a 100 g solution:
- Mass of NH₃ = 30 g
- Mass of water = 100 g - 30 g = 70 g

Now, we need to find out how much NH₃ is dissolved in 105 g of water.

Using the ratio:
\[ \text{Mass of NH₃ in 105 g of water} = \left(\frac{30 \text{ g NH₃}}{70 \text{ g water}}\right) \times 105 \text{ g water} = 45 \text{ g NH₃} \]

### Step 2: Calculate the mass of the saturated solution
The total mass of the saturated solution can be calculated by adding the mass of water and the mass of NH₃.

\[ \text{Mass of saturated solution} = \text{Mass of water} + \text{Mass of NH₃} = 105 \text{ g} + 45 \text{ g} = 150 \text{ g} \]

### Step 3: Calculate the volume of the saturated solution
Using the density of the solution, we can find the volume of the saturated solution.

Given:
- Density of solution = 0.9 g/mL

Using the formula:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{150 \text{ g}}{0.9 \text{ g/mL}} = 166.67 \text{ mL} \]

### Step 4: Calculate the number of moles of NH₃
Now, we will calculate the number of moles of NH₃ using its mass and molecular weight.

Given:
- Mass of NH₃ = 45 g
- Molar mass of NH₃ = 17 g/mol

Using the formula:
\[ \text{Number of moles} (n) = \frac{\text{Mass}}{\text{Molar mass}} = \frac{45 \text{ g}}{17 \text{ g/mol}} \approx 2.65 \text{ mol} \]

### Step 5: Calculate the volume of NH₃ gas at given conditions
We will use the ideal gas law to find the volume of NH₃ gas at 4°C and 775 mm Hg.

First, convert the temperature to Kelvin:
\[ T = 4 + 273.15 = 277.15 \text{ K} \]

Convert the pressure from mm Hg to atm:
\[ P = \frac{775 \text{ mm Hg}}{760 \text{ mm Hg/atm}} \approx 1.02 \text{ atm} \]

Using the ideal gas equation \(PV = nRT\):
\[ V = \frac{nRT}{P} \]
Where:
- R = 0.0821 L·atm/(K·mol)

Substituting the values:
\[ V = \frac{(2.65 \text{ mol}) \times (0.0821 \text{ L·atm/(K·mol)}) \times (277.15 \text{ K})}{1.02 \text{ atm}} \approx 59.03 \text{ L} \]

### Final Answer
The volume of NH₃ gas at 4°C and 775 mm Hg is approximately **59.03 liters**.

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