COMEDK · Chemistry · 17. Solutions
0.60 g of urea is dissolved in 360 g of water at room temperature. The vapour pressure of pure water is 35 mm of Hg , the lowering of vapour pressure at this temperature is:
[Molar mass of urea is \(60 \mathrm{~g} \mathrm{~mol}^{-1}\) ]
- A 0.025 mm of Hg
- B 0.17 mm of Hg
- C 0.25 mm of Hg
- D 0.017 mm of Hg
Answer & Solution
Correct Answer
(D) 0.017 mm of Hg
Step-by-step Solution
Detailed explanation
The molar mass of urea is \(M_{urea} = 60 \text{ g mol}^{-1}\). The mass of urea is \(w_{urea} = 0.60 \text{ g}\). The number of moles of urea is \(n_{urea} = \dfrac{0.60}{60} = 0.01 \text{ mol}\). The mass of water is \(w_{water} = 360 \text{ g}\). The molar mass of water is…
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