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Vapour pressure of benzene at 30^(@)C is...

Vapour pressure of benzene at `30^(@)C` is 121.8 mm. When 15 g of a non-volatile solute is dissolved in 250 g of benzene its vapour pressure decreased to 120.2 mm. The molecular weight of the solute is (mol. Weight of solvent = 78)

A

`356.2`

B

`456.8`

C

`530.1`

D

`656.7`

Text Solution

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The correct Answer is:
To find the molecular weight of the non-volatile solute, we can use Raoult's Law, which relates the vapor pressure of a solvent to the concentration of a solute in that solvent. The formula we will use is: \[ \frac{P_0 - P_s}{P_0} = \frac{w_s \cdot M_b}{M_s \cdot w_b} \] Where: - \(P_0\) = vapor pressure of pure solvent (benzene) = 121.8 mm - \(P_s\) = vapor pressure of the solution = 120.2 mm - \(w_s\) = weight of the solute = 15 g - \(M_b\) = molecular weight of the solvent (benzene) = 78 g/mol - \(M_s\) = molecular weight of the solute (unknown) - \(w_b\) = weight of the solvent (benzene) = 250 g ### Step 1: Calculate the change in vapor pressure First, we need to find the change in vapor pressure: \[ P_0 - P_s = 121.8 \, \text{mm} - 120.2 \, \text{mm} = 1.6 \, \text{mm} \] ### Step 2: Substitute the values into the equation Now we can substitute the values into the equation: \[ \frac{1.6}{121.8} = \frac{15 \cdot 78}{M_s \cdot 250} \] ### Step 3: Simplify the left side Calculate the left side: \[ \frac{1.6}{121.8} \approx 0.0131 \] ### Step 4: Set up the equation Now we can set up the equation: \[ 0.0131 = \frac{15 \cdot 78}{M_s \cdot 250} \] ### Step 5: Calculate the numerator Calculate the numerator: \[ 15 \cdot 78 = 1170 \] ### Step 6: Substitute back into the equation Now substitute back into the equation: \[ 0.0131 = \frac{1170}{M_s \cdot 250} \] ### Step 7: Rearrange to solve for \(M_s\) Rearranging gives us: \[ M_s \cdot 250 \cdot 0.0131 = 1170 \] \[ M_s = \frac{1170}{250 \cdot 0.0131} \] ### Step 8: Calculate \(M_s\) Now calculate \(M_s\): \[ M_s = \frac{1170}{3.275} \approx 356.2 \, \text{g/mol} \] ### Conclusion The molecular weight of the non-volatile solute is approximately **356.2 g/mol**.
To find the molecular weight of the non-volatile solute, we can use Raoult's Law, which relates the vapor pressure of a solvent to the concentration of a solute in that solvent. The formula we will use is: \[ \frac{P_0 - P_s}{P_0} = \frac{w_s \cdot M_b}{M_s \cdot w_b} \] Where: - \(P_0\) = vapor pressure of pure solvent (benzene) = 121.8 mm ...
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