0.6 g of a gas at `15^(@)C` and 745 mm Hg occupies `200cm^(3)`. It occupies `182.6cm^(3)` in dry state at N.T.P. Calculate the aqueous tension at `15^(@)C.`

Explanation

To solve the problem of calculating the aqueous tension at 15°C, we will follow these steps:

### Step 1: Understand the given data
We have the following information:
- Mass of gas (m) = 0.6 g
- Temperature (T1) = 15°C = 15 + 273 = 288 K
- Pressure (P1) = 745 mm Hg
- Volume of gas at P1 and T1 (V1) = 200 cm³
- Volume of gas at NTP (V2) = 182.6 cm³
- Pressure at NTP (P2) = 760 mm Hg
- Temperature at NTP (T2) = 20°C = 20 + 273 = 293 K

### Step 2: Set up the equation
We will use the ideal gas law in the form of the combined gas law:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Where:
- \(P_1\) is the total pressure of the moist gas,
- \(V_1\) is the volume of the moist gas,
- \(T_1\) is the temperature in Kelvin,
- \(P_2\) is the pressure of the dry gas at NTP,
- \(V_2\) is the volume of the dry gas,
- \(T_2\) is the temperature at NTP.

### Step 3: Express \(P_1\) in terms of aqueous tension
Let \(x\) be the aqueous tension at 15°C. The pressure of the dry gas can be expressed as:
\[ P_{\text{dry}} = P_1 - x = 745 - x \]
Now we can substitute this into the combined gas law equation.

### Step 4: Substitute values into the equation
Substituting the known values into the equation:
\[ \frac{(745 - x) \cdot 200}{288} = \frac{760 \cdot 182.6}{293} \]

### Step 5: Solve for \(x\)
First, calculate the right-hand side:
\[ \frac{760 \cdot 182.6}{293} \approx 482.04 \]
Now, we can set up the equation:
\[ (745 - x) \cdot 200 = 482.04 \cdot 288 \]
Calculating the right side:
\[ 482.04 \cdot 288 \approx 138,336.32 \]
Now, we have:
\[ (745 - x) \cdot 200 = 138336.32 \]
Dividing both sides by 200:
\[ 745 - x = \frac{138336.32}{200} \approx 691.68 \]
Now, solving for \(x\):
\[ x = 745 - 691.68 \approx 53.32 \text{ mm Hg} \]

### Step 6: Conclusion
The aqueous tension at 15°C is approximately **53.32 mm Hg**.