AP EAMCET · Chemistry · Solutions
At \(300 \mathrm{~K}\) an ideal solution is formed by mixing \(460 \mathrm{~g}\) of toluene with \(390 \mathrm{~g}\) benzene. If the vapour pressure of pure toluene and benzene at \(300 \mathrm{~K}\) are 32 and \(40 \mathrm{~mm}\) respectively, the mole fraction of toluene in vapour phase is
- A 0.196
- B 0.588
- C 0.294
- D 0.444
Answer & Solution
Correct Answer
(D) 0.444
Step-by-step Solution
Detailed explanation
Given, \(w_a=460 \mathrm{~g}\) (toluene) \(w_B=390 \mathrm{~g} \text { (benzene) }\) Vapour pressure of pure toluene \(\left(p_A\right)=32 \mathrm{~mm}\) Vapour pressure of pure benzene \(\left(p_B\right)=40 \mathrm{~mm}\) Moles of toluene,…
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