JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
What weight of glucose must be dissolved in \(100 g\) of water to lower the vapour pressure by \(0.20\, mm\) \(Hg\) ? (Assume dilute solution is being formed) Given: Vapour pressure of pure water is \(54.2\,mm\) \(Hg\) at room temperature. Molar mass of glucose is \(180\,g\,mol ^{-1}\)
- A \(4.69\,g\)
- B \(3.59\,g\)
- C \(2.59\,g\)
- D \(3.69\,g\)
Answer & Solution
Correct Answer
(D) \(3.69\,g\)
Step-by-step Solution
Detailed explanation
\(\frac{ P ^0- P _5}{ P ^0}=\frac{ n }{ N }\) (for dilute solution) \(\frac{0.2}{54.2}=\frac{ n \times 18}{100}\) \(n =\frac{100}{271 \times 18}\) \(w =\frac{100 \times 180}{271 \times 18} ; w =3.69\,g\)
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