KCET · Chemistry · Solutions
180 g of glucose, \(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6\), is dissolved in 1 kg of water in a vessel. The temperature at which water boils at 1.013 bar is ______ (given, \(\mathrm{K}_{\mathrm{b}}\) for water is \(052 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\). Boiling point for pure water is 373.15 K )
- A 373.67 K
- B 373015 K
- C 373.0 K
- D 373.202 K
Answer & Solution
Correct Answer
(A) 373.67 K
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \Delta \mathrm{T}_{\mathrm{b}}=\mathrm{i} \mathrm{K}_{\mathrm{b}} \mathrm{m} \quad(\mathrm{i}=1) \text { glowse } \\ & \mathrm{T}_{\mathrm{b}}-373.15=1 \times 0.52 \times \frac{180}{180 \times 1} \\ & \mathrm{~T}_{\mathrm{b}}=0.52+373.15=373.67 \mathrm{~K}\end{aligned}\)
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