6 litre `H_(2)O` is placed in a closed evacuated room of volume 8.27 litre at the temperature 300K. The density of liquid water at 300 K is 1.0 gm/ml. the vapour pressure of water at 300 K is 22.8 mm Hg. Neglect the change in volume of liquid water by vaporization. `{:("List I",,"List II"),("(P) Mass of watwer vapour formed (in gm)",,"(1) 6"),("(Q) Moles of water vapour fomed",,"(2) 18"),("(R) Approx. mass of liquid water left (in kg)",,"(3) 3"),("(S) Total moles of atoms in vapour form",,"(4) 1"):}`

Options

A. `{:(P,Q,R,S),(1,2,4,3):}`
B. `{:(P,Q,R,S),(4,3,2,1):}`
C. `{:(P,Q,R,S),(2,4,1,3):}`
D. `{:(P,Q,R,S),(1,2,3,4):}`

Answer

To solve the problem step-by-step, we will follow these calculations:

### Step 1: Convert the vapor pressure from mm Hg to atm
The vapor pressure of water at 300 K is given as 22.8 mm Hg. We need to convert this to atm using the conversion factor \(1 \text{ atm} = 760 \text{ mm Hg}\).

\[ P = \frac{22.8 \text{ mm Hg}}{760 \text{ mm Hg/atm}} = 0.03 \text{ atm} \]

### Step 2: Use the ideal gas law to find the number of moles of water vapor
The ideal gas equation is given by:

\[ PV = nRT \]

Where:
- \(P\) = pressure in atm
- \(V\) = volume in liters (8.27 L)
- \(n\) = number of moles
- \(R\) = ideal gas constant (0.0821 L·atm/(K·mol))
- \(T\) = temperature in Kelvin (300 K)

Rearranging the equation to solve for \(n\):

\[ n = \frac{PV}{RT} \]

Substituting the values:

\[ n = \frac{(0.03 \text{ atm})(8.27 \text{ L})}{(0.0821 \text{ L·atm/(K·mol)})(300 \text{ K})} \]

Calculating \(n\):

\[ n = \frac{0.2481}{24.63} \approx 0.0101 \text{ moles} \]

### Step 3: Calculate the mass of water vapor formed
The molar mass of water (H₂O) is approximately 18 g/mol. To find the mass of the water vapor formed, we use the formula:

\[ \text{Mass} = n \times \text{Molar mass} \]

Substituting the values:

\[ \text{Mass} = 0.0101 \text{ moles} \times 18 \text{ g/mol} \approx 0.182 \text{ g} \]

### Step 4: Calculate the mass of liquid water initially present
The volume of liquid water is given as 6 liters. Since the density of water is 1.0 g/mL, we convert liters to mL:

\[ 6 \text{ L} = 6000 \text{ mL} \]

Now, using the density to find the mass:

\[ \text{Mass} = \text{Density} \times \text{Volume} = 1.0 \text{ g/mL} \times 6000 \text{ mL} = 6000 \text{ g} = 6 \text{ kg} \]

### Step 5: Calculate the total moles of atoms in vapor form
Each water molecule (H₂O) consists of 3 atoms (2 H and 1 O). Therefore, the total moles of atoms in the vapor form can be calculated as:

\[ \text{Total moles of atoms} = n \times 3 \]

Substituting the value of \(n\):

\[ \text{Total moles of atoms} = 0.0101 \text{ moles} \times 3 \approx 0.0303 \text{ moles} \]

### Summary of Results
- (P) Mass of water vapor formed: **0.182 g**
- (Q) Moles of water vapor formed: **0.0101 moles**
- (R) Approx. mass of liquid water left: **6 kg**
- (S) Total moles of atoms in vapor form: **0.0303 moles**

### Matching with Lists
- (P) Mass of water vapor formed (in gm): **0.182**
- (Q) Moles of water vapor formed: **0.0101**
- (R) Approx. mass of liquid water left (in kg): **6**
- (S) Total moles of atoms in vapor form: **0.0303**