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JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties

\(40\, \mathrm{~g}\) of glucose (Molar mass \(=180\) ) is mixed with \(200\, \mathrm{~mL}\) of water. The freezing point of solution is \(.....\,\mathrm{K}\). (Nearest integer) [Given : \(\mathrm{K}_{\mathrm{f}}=1.86 \,\mathrm{~K} \,\mathrm{~kg} \,\mathrm{~mol}^{-1} ;\) Density of water \(=\) \(1.00 \,\mathrm{~g}\, \mathrm{~cm}^{-3} ;\) Freezing point of water \(\left.=273.15\, \mathrm{~K}\right]\)

  1. A \(271\)
  2. B \(370\)
  3. C \(71\)
  4. D \(521\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(271\)

Step-by-step Solution

Detailed explanation

\(\text { molality }=\frac{\left(\frac{40}{180}\right) \,{\text {mol }}}{0.2 \,\mathrm{Kg}}=\left(\frac{10}{9}\right) \,\text { molal }\) \(\Rightarrow \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{T}_{\mathrm{f}}-\mathrm{T}_{\mathrm{f}}^{\prime}=1.86 \times \frac{10}{9}\)…
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