A solution containing 20.0 grams of a nonvolatile solute in exactly 1.00 mole of volatile solvent has a vapoure pressure of 0.500 atm at `20^(@)C`. A second mole of solvet is added to the mixture, and the resulting solution has a vapoure pressure of `0.550` at of `20^(@)C`. What is the molecular weight of the solute ?( g/mole)

Options

A. `45.0`
B. `90.0`
C. `20.0`
D. `50.0`

Answer

To find the molecular weight of the non-volatile solute, we can follow these steps:

### Step 1: Define Variables
Let:
- \( P \) = vapor pressure of the pure solvent at \( 20^\circ C \)
- \( P_1 = 0.500 \, \text{atm} \) (vapor pressure of the solution with 1 mole of solvent)
- \( P_2 = 0.550 \, \text{atm} \) (vapor pressure of the solution with 2 moles of solvent)
- \( n_s = 1 \, \text{mol} \) (initial moles of solvent)
- \( n_s' = 2 \, \text{mol} \) (final moles of solvent)
- \( n_{solute} = \frac{20 \, \text{g}}{M} \) (moles of solute, where \( M \) is the molar mass of the solute)

### Step 2: Apply Raoult's Law
Using Raoult's Law, we can express the change in vapor pressure for both scenarios:

1. For the first solution:
\[ \frac{P - P_1}{P} = \frac{n_{solute}}{n_s} \]
Substituting values:
\[ \frac{P - 0.500}{P} = \frac{\frac{20}{M}}{1} \]
Rearranging gives:
\[ P - 0.500 = \frac{20P}{M} \]
This can be rewritten as:
\[ PM - 0.500M = 20P \quad \text{(Equation 1)} \]

2. For the second solution:
\[ \frac{P - P_2}{P} = \frac{n_{solute}}{n_s'} \]
Substituting values:
\[ \frac{P - 0.550}{P} = \frac{\frac{20}{M}}{2} \]
Rearranging gives:
\[ P - 0.550 = \frac{10P}{M} \]
This can be rewritten as:
\[ PM - 0.550M = 10P \quad \text{(Equation 2)} \]

### Step 3: Solve the Equations
Now we have two equations:
1. \( PM - 0.500M = 20P \)
2. \( PM - 0.550M = 10P \)

Subtract Equation 2 from Equation 1:
\[ (PM - 0.500M) - (PM - 0.550M) = 20P - 10P \]
This simplifies to:
\[ 0.050M = 10P \]
Thus:
\[ M = \frac{10P}{0.050} = 200P \]

### Step 4: Substitute \( P \) to Find \( M \)
We need to find \( P \). We can use either equation to find \( P \). Let's use Equation 1:
\[ PM - 0.500M = 20P \]
Substituting \( M = 200P \):
\[ P(200P) - 0.500(200P) = 20P \]
This simplifies to:
\[ 200P^2 - 100P = 20P \]
Rearranging gives:
\[ 200P^2 - 120P = 0 \]
Factoring out \( P \):
\[ P(200P - 120) = 0 \]
Thus:
\[ 200P - 120 = 0 \implies P = \frac{120}{200} = 0.6 \, \text{atm} \]

### Step 5: Calculate Molar Mass
Substituting \( P \) back into \( M \):
\[ M = 200 \times 0.6 = 120 \, \text{g/mol} \]

### Final Answer
The molecular weight of the solute is **120 g/mol**.

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