AP EAMCET · Chemistry · Solutions
A non-volatile solute is dissolved in water. The \(\Delta T_b\) of resultant solution is 0.052 K . What is the freezing point of the solution (in K)?
( \(\mathrm{K}_{\mathrm{b}}\) of water \(=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} ; \mathrm{K}_{\mathrm{f}}\) of water \(=1.86 \mathrm{~K} \mathrm{~kg}\) \(\mathrm{mol}^{-1}\), Freezing point of water \(=273 \mathrm{~K}\) )
- A \(272.628\)
- B \(273.186\)
- C \(273.000\)
- D \(272.814\)
Answer & Solution
Correct Answer
(D) \(272.814\)
Step-by-step Solution
Detailed explanation
Given \(\begin{aligned} & \Delta \mathrm{T}_{\mathrm{b}}=0.052 \mathrm{~K} \\ & 0.052 \mathrm{~K}=0.52 \times \mathrm{m} \\ & \mathrm{~m}=0.1 \\ & \Delta \mathrm{~T}_{\mathrm{f}}=\mathrm{m} \mathrm{~K}_{\mathrm{f}}=0.1 \times 1.86=0.186 \end{aligned}\) Freezing Point of solution…
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