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Question: Answered & Verified by Expert
At $\mathrm{T}(\mathrm{K})$, the vapour pressure of pure benzene (molar mass $=78 \mathrm{~g} \mathrm{~mol}^{-1}$ ) is $0.85 \mathrm{bar}$. When $2.0 \mathrm{~g}$ of non-volatile, non-electrolyte solute is added to $39 \mathrm{~g}$ of benzene, the vapour pressure of solution at $\mathrm{T}(\mathrm{K})$ is 0.83 bar. The elevation in boiling point (in $\mathrm{K}$ ) of the same solution is: $\left(\mathrm{K}_{\mathrm{b}}\right.$ of benzene is $2.6 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$)
ChemistrySolutionsAP EAMCETAP EAMCET 2017 (25 Apr Shift 2)
Options:
  • A $0.0784$
  • B $0.196$
  • C $1.568$
  • D $0.784$
Solution:
2021 Upvotes Verified Answer
The correct answer is: $0.784$
No solution. Refer to answer key.

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