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Solutions question

2005 · Shift 0 · Q18
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Solutions question

2005 · Shift 0 · Q18

JEE MainChemistrySolutionsMCQ+4 / −1
Benzene and toluene form nearly ideal solutions. At 20 oC, the vapour pressure of benzene is 75 torr and that of toluene is 22 torr. The partial vapour pressure of benzene at 20 oC for a solution containing 78 g of benzene and 46 g of toluene in torr is
  1. A
    50
  2. B
    25
  3. C
    53.5
  4. D
    37.5
View written solutionFree

Correct answer: A

  1. Use Raoult’s law for an ideal solution

For benzene in an ideal solution, pbenzene=xbenzene pbenzene∘p_{\text{benzene}} = x_{\text{benzene}}\, p^\circ_{\text{benzene}}pbenzene​=xbenzene​pbenzene∘​ where:

  • xbenzenex_{\text{benzene}}xbenzene​ = mole fraction of benzene in solution
  • pbenzene∘=75 torrp^\circ_{\text{benzene}} = 75\ \text{torr}pbenzene∘​=75 torr
  1. Calculate moles of each component
  • Molar mass of benzene (C6H6)=78 g mol−1\left(C_6H_6\right) = 78\ \text{g mol}^{-1}(C6​H6​)=78 g mol−1
  • Molar mass of toluene (C7H8)=92 g mol−1\left(C_7H_8\right) = 92\ \text{g mol}^{-1}(C7​H8​)=92 g mol−1

So, nbenzene=7878=1 moln_{\text{benzene}} = \frac{78}{78} = 1\ \text{mol}nbenzene​=7878​=1 mol ntoluene=4692=0.5 moln_{\text{toluene}} = \frac{46}{92} = 0.5\ \text{mol}ntoluene​=9246​=0.5 mol

  1. Find mole fraction of benzene
= \frac{1}{1+0.5} = \frac{1}{1.5} = \frac{2}{3}$$ 4. **Calculate partial vapour pressure of benzene** $$p_{\text{benzene}} = x_{\text{benzene}}\, p^\circ_{\text{benzene}} = \frac{2}{3} \times 75 = 50\ \text{torr}$$ 5. **Check options** - A: $50$ ✅ - B: $25$ ❌ - C: $53.5$ ❌ - D: $37.5$ ❌ Therefore, the correct answer is **A**.
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