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64 g non - volatile solute is added to 7...

64 g non - volatile solute is added to 702 g benzene. The vapour pressure of benzene has decreased from 200 mm of Hg to 180 mm of Hg. Molecular weight of the solute is

A

128

B

64

C

96

D

256

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we will use Raoult's Law and the concept of mole fraction. Here’s a step-by-step breakdown of the solution: ### Step 1: Identify the given data - Mass of non-volatile solute (Wb) = 64 g - Mass of benzene (Wa) = 702 g - Vapor pressure of pure benzene (P0a) = 200 mm Hg - Vapor pressure of the solution (Pa) = 180 mm Hg ### Step 2: Calculate the change in vapor pressure Using Raoult's Law, we can find the decrease in vapor pressure: \[ \Delta P = P_0 - P = 200 \, \text{mm Hg} - 180 \, \text{mm Hg} = 20 \, \text{mm Hg} \] ### Step 3: Calculate the mole fraction of the solute (XB) Using the formula from Raoult's Law: \[ \frac{P_0 - P}{P_0} = X_B \] Substituting the values: \[ X_B = \frac{20}{200} = \frac{1}{10} \] ### Step 4: Relate mole fraction to moles The mole fraction of the solute (XB) can also be expressed as: \[ X_B = \frac{N_B}{N_A + N_B} \] Where: - \(N_A\) = number of moles of benzene - \(N_B\) = number of moles of solute ### Step 5: Calculate the number of moles of benzene (NA) The molecular weight of benzene (C6H6) is: \[ \text{Molar mass of C6H6} = (12 \times 6) + (1 \times 6) = 72 + 6 = 78 \, \text{g/mol} \] Now, calculate the number of moles of benzene: \[ N_A = \frac{W_a}{\text{Molar mass of benzene}} = \frac{702 \, \text{g}}{78 \, \text{g/mol}} = 9 \, \text{moles} \] ### Step 6: Substitute into the mole fraction equation Substituting \(N_A\) into the mole fraction equation: \[ \frac{1}{10} = \frac{N_B}{9 + N_B} \] ### Step 7: Solve for NB Cross-multiplying gives: \[ N_B = \frac{1}{10} (9 + N_B) \] Multiplying through by 10: \[ 10N_B = 9 + N_B \] Rearranging gives: \[ 10N_B - N_B = 9 \implies 9N_B = 9 \implies N_B = 1 \, \text{mole} \] ### Step 8: Calculate the molecular weight of the solute Using the formula for moles: \[ N_B = \frac{W_b}{M_B} \] Where \(M_B\) is the molecular weight of the solute. Rearranging gives: \[ M_B = \frac{W_b}{N_B} = \frac{64 \, \text{g}}{1 \, \text{mole}} = 64 \, \text{g/mol} \] ### Final Answer The molecular weight of the solute is **64 g/mol**. ---
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