`105 mL` of pure water at `4^(circ)C` saturated with `NH_(3)` gas yielded a solution of density `0.9g mL^(-1)` and containing `30%NH_(3)` by mass. Find out the volume of `NH_(3)` solution resulting and the volume of `NH_(3)` gas at `4^(circ)C` and `775 mm` of `Hg`, which was used to saturate water.

Options

A. 66.67 ml
B. 166.67 ml
C. 133.33 ml
D. 266.67 ml

Explanation

To solve the problem step by step, we will follow the instructions provided in the video transcript and perform the necessary calculations.

### Step 1: Calculate the mass of the solution
We know the density of the solution is `0.9 g/mL`. Let \( V \) be the volume of the solution in mL. The mass of the solution can be calculated using the formula:

\[ \text{Mass of solution} = \text{Volume} \times \text{Density} \]

Thus,

\[ \text{Mass of solution} = 0.9 \, V \, \text{g} \]

### Step 2: Calculate the mass of water
The volume of water is given as `105 mL`. The density of pure water is approximately `1 g/mL`, so the mass of water is:

\[ \text{Mass of water} = \text{Volume} \times \text{Density} = 105 \, \text{mL} \times 1 \, \text{g/mL} = 105 \, \text{g} \]

### Step 3: Determine the mass of ammonia in the solution
The solution is given to contain `30% NH₃` by mass. This means that `70%` of the solution is water. Therefore, we can express the mass of water in the solution as:

\[ \text{Mass of water in solution} = \text{Mass of solution} \times \frac{70}{100} = 0.9 \, V \times 0.7 = 0.63 \, V \, \text{g} \]

### Step 4: Set up the equation
Now, we equate the mass of water in the solution to the mass of water we calculated earlier:

\[ 0.63 \, V = 105 \]

### Step 5: Solve for \( V \)
To find \( V \), we rearrange the equation:

\[ V = \frac{105}{0.63} \]

Calculating this gives:

\[ V \approx 166.67 \, \text{mL} \]

### Step 6: Calculate the volume of NH₃ gas
Next, we need to find the volume of ammonia gas at `4°C` and `775 mm Hg`.

Using the ideal gas law, we can find the volume of the gas. First, we need to find the mass of ammonia in the solution:

\[ \text{Mass of NH₃} = \text{Mass of solution} \times \frac{30}{100} = 0.9 \, V \times 0.3 = 0.27 \, V \, \text{g} \]

Substituting \( V = 166.67 \, \text{mL} \):

\[ \text{Mass of NH₃} = 0.27 \times 166.67 \approx 45 \, \text{g} \]

Now, we convert this mass to moles:

\[ \text{Molar mass of NH₃} = 14 + (3 \times 1) = 17 \, \text{g/mol} \]

\[ \text{Moles of NH₃} = \frac{45}{17} \approx 2.65 \, \text{mol} \]

Using the ideal gas law \( PV = nRT \) to find the volume of NH₃ gas:

- \( P = 775 \, \text{mm Hg} = \frac{775}{760} \, \text{atm} \approx 1.02 \, \text{atm} \)
- \( R = 0.0821 \, \text{L atm/(K mol)} \)
- \( T = 4°C = 277 \, \text{K} \)

Now substituting into the ideal gas equation:

\[ V = \frac{nRT}{P} = \frac{2.65 \times 0.0821 \times 277}{1.02} \]

Calculating this gives:

\[ V \approx 63.5 \, \text{L} \]

### Final Answers
- Volume of NH₃ solution = **166.67 mL**
- Volume of NH₃ gas = **63.5 L**