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The vapour pressure of pure benzene at a certain temperature is 640 mm Hg. A non-volatile, non-electrolyte solid weighing 2.175 g is added to 39.0 g of benzene. The vapour pressure of the solution is 600 mm Hg. Molecular weight of the non-electrolyte is 
  • a)
    65.25 g mol-1
  • b)
    60.08 g mol-1
  • c)
    72.08 g mol-1
  • d)
    92.05 g mol-1
Correct answer is option 'A'. Can you explain this answer?
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The vapour pressure of pure benzene at a certain temperature is 640 mm...
 
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The vapour pressure of pure benzene at a certain temperature is 640 mm...
Given:
Vapour pressure of pure benzene (P1) = 640 mm Hg
Weight of non-volatile, non-electrolyte solid (W2) = 2.175 g
Weight of benzene (W1) = 39.0 g
Vapour pressure of the solution (P2) = 600 mm Hg

To find:
Molecular weight of the non-electrolyte (M2)

Formula Used:
Raoult's Law: P2 = P1*X1
Where P2 is the vapour pressure of the solution, P1 is the vapour pressure of the pure solvent, and X1 is the mole fraction of the solvent in the solution.

Solution:
Step 1: Calculate the mole fraction of benzene in the solution
Mole fraction of benzene (X1) = (moles of benzene)/(total moles)
To calculate the moles of benzene, we need to know the number of moles of benzene and the number of moles of the non-electrolyte.

Step 2: Calculate the moles of benzene
Moles of benzene (n1) = (weight of benzene)/(molar mass of benzene)
The molar mass of benzene (C6H6) = 12.01 g/mol (carbon) + 1.008 g/mol (hydrogen) x 6 = 78.11 g/mol

Plugging in the values:
n1 = 39.0 g / 78.11 g/mol = 0.499 mol

Step 3: Calculate the moles of the non-electrolyte
Moles of the non-electrolyte (n2) = (weight of the non-electrolyte)/(molar mass of the non-electrolyte)
Plugging in the values:
n2 = 2.175 g / M2 g/mol

Step 4: Calculate the mole fraction of benzene
X1 = n1/(n1 + n2)
Substituting the values:
X1 = 0.499 / (0.499 + 2.175/M2)

Step 5: Apply Raoult's Law to find the molecular weight of the non-electrolyte
P2 = P1 * X1
Substituting the values:
600 mm Hg = 640 mm Hg * (0.499 / (0.499 + 2.175/M2))

Simplifying the equation:
600 = 640 * (0.499 / (0.499 + 2.175/M2))

To solve this equation, we can cross-multiply and solve for M2:

600 * (0.499 + 2.175/M2) = 640 * 0.499
299.4 + 1305/M2 = 319.36
1305/M2 = 319.36 - 299.4
1305/M2 = 19.96
M2 = 1305/19.96
M2 = 65.25 g/mol

Therefore, the molecular weight of the non-electrolyte is 65.25 g/mol
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Vapour pressure of a solvent is the pressure exerted by vapours when they are in equilibrium with its solvent at that temperature. The vapour pressure of solvent is dependent on nature of solvent, temperature, addition of non-volatile solute as well as nature of solute to dissociate or associate.The vapurpressuere ofa mixture obtained by mixing two volatile liquids is given by PM = . XA + . XBwhereandare vapour pressures of pure components A and B and XA, XBare their mole fraction in mixture . For solute-solvent system, the realtion become PM = . XAwhere B is non-volatile solute.Q. A mixture of two volatile liquids A and B for 1 and 3 moles respectivley has a V.P. of 300 mm at 27C. If one mole of A is further added to this solution, the vapour pressure becomes 290 mm at 27C. The vapour pressure of pure A is

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The vapour pressure of pure benzene at a certain temperature is 640 mm Hg. A non-volatile, non-electrolyte solid weighing 2.175 g is added to 39.0 g of benzene. The vapour pressure of the solution is 600 mm Hg. Molecular weight of the non-electrolyte isa)65.25 g mol-1b)60.08g mol-1c)72.08g mol-1d)92.05g mol-1Correct answer is option 'A'. Can you explain this answer?
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