MHT CET · Chemistry · Solutions
Identify the correct relation between depression in freezing point and freezing point of pure solvent?
- A \(\mathrm{T}^{\circ}=\mathrm{T} \times \Delta \mathrm{T}_{\mathrm{f}}\)
- B \(\mathrm{T}^{\circ}=\Delta \mathrm{T}_{\mathrm{f}}-\mathrm{T}\)
- C \(\mathrm{T}^{\circ}=\mathrm{T}-\Delta \mathrm{T}_{\mathrm{f}}\)
- D \(\mathrm{T}^{\circ}=\Delta \mathrm{T}_{\mathrm{f}}+\mathrm{T}\)
Answer & Solution
Correct Answer
(D) \(\mathrm{T}^{\circ}=\Delta \mathrm{T}_{\mathrm{f}}+\mathrm{T}\)
Step-by-step Solution
Detailed explanation
\(\Delta T_{f}=T^{0}-T\)
\(\therefore T^{0}=\Delta T_{f}+T\)
\(T^{0}=\) freezing point of pure solvent
\(T=\) freezing point of solution
\(\Delta T_{f}=\) depression in freezing point
\(\therefore T^{0}=\Delta T_{f}+T\)
\(T^{0}=\) freezing point of pure solvent
\(T=\) freezing point of solution
\(\Delta T_{f}=\) depression in freezing point
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