A solution containing 28 g of phosphorus in 315 g `CS_(2)(b.p. 46.3^(@)C`) boils at `47.98^(@)C` . If `K_(b)` for `CS_(2)` is `2.34` K kg `mol^(-1)` . The formula of phosphorus is (at , mass of P = 31).

Options

A. `P_(4)`
B. `P_(8)`
C. `P_(2)`
D. None

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the elevation in boiling point (ΔTb)
The elevation in boiling point can be calculated using the formula:
\[ \Delta T_b = T_b(\text{solution}) - T_b(\text{solvent}) \]
Given:
- \(T_b(\text{solution}) = 47.98^\circ C\)
- \(T_b(\text{solvent}) = 46.3^\circ C\)

Calculating ΔTb:
\[ \Delta T_b = 47.98 - 46.3 = 1.68^\circ C \]

### Step 2: Use the boiling point elevation formula
The boiling point elevation is also given by the formula:
\[ \Delta T_b = K_b \cdot m \]
where:
- \(K_b\) is the ebullioscopic constant of the solvent (CS2), given as \(2.34 \, \text{K kg mol}^{-1}\)
- \(m\) is the molality of the solution

### Step 3: Rearrange the formula to find molality (m)
From the boiling point elevation formula, we can rearrange it to find molality:
\[ m = \frac{\Delta T_b}{K_b} \]

Substituting the values:
\[ m = \frac{1.68}{2.34} \approx 0.717 \, \text{mol/kg} \]

### Step 4: Calculate the number of moles of solute (phosphorus)
Molality is defined as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} \]
We need to convert the mass of the solvent from grams to kilograms:
\[ \text{mass of solvent} = 315 \, \text{g} = 0.315 \, \text{kg} \]

Now we can find the moles of solute:
\[ \text{moles of solute} = m \cdot \text{mass of solvent (kg)} = 0.717 \cdot 0.315 \approx 0.226 \, \text{mol} \]

### Step 5: Calculate the molar mass of phosphorus
We know the mass of phosphorus used in the solution is 28 g. To find the molar mass (M) of phosphorus, we can use the formula:
\[ \text{Molar mass} = \frac{\text{mass of solute (g)}}{\text{moles of solute}} \]
Substituting the values:
\[ \text{Molar mass} = \frac{28 \, \text{g}}{0.226 \, \text{mol}} \approx 123.88 \, \text{g/mol} \]

### Step 6: Determine the formula of phosphorus
Given that the atomic mass of phosphorus (P) is 31 g/mol, we can determine the number of phosphorus atoms in the molecular formula:
\[ \text{Number of phosphorus atoms} = \frac{123.88 \, \text{g/mol}}{31 \, \text{g/mol}} \approx 4 \]
Thus, the molecular formula of phosphorus is \(P_4\).

### Final Answer
The formula of phosphorus is \(P_4\).

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