AP EAMCET · Chemistry · Solutions
The vapour pressure of water at \(23^{\circ} \mathrm{C}\) is \(19.8 \mathrm{~mm}, 0.1\) mole of glucose is dissolved in \(178.2 \mathrm{~g}\) of water. What is the vapour pressure (in \(\mathrm{mm}\) ) of the resultant solution?
- A 19
- B 19.602
- C 19.402
- D 19.202
Answer & Solution
Correct Answer
(B) 19.602
Step-by-step Solution
Detailed explanation
Given \(P_s=19.8 \mathrm{~mm}\) \(\begin{aligned} & n_A=0.1 \\ & n_B=\frac{178.2}{18}=9.9 \end{aligned}\) According to Raoult's law \(\begin{gathered} \frac{P_s-P}{P_s}=\frac{n_A}{n_A+n_B} \\ \frac{19.8-P}{19.8}=\frac{0.1}{9.9+0.1} \end{gathered}\) or…
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