JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
\(1.80\,g\) of solute \(A\) was dissolved in \(62.5\,cm ^{3}\) of ethanol and freezing point of the solution was found to be \(155.1\,K\). The molar mass of solute \(A\) is \(..........g\,\,mol ^{-1}\). [Given: Freezing point of ethanol is \(156.0 K\) Density of ethanol is \(0.80\,g\,cm ^{-3}\). Freezing point depression constant of ethanol is \(\left.2.00\,K\,kg\,mol\,m ^{-1}\right]\)
- A \(81\)
- B \(80\)
- C \(82\)
- D \(83\)
Answer & Solution
Correct Answer
(B) \(80\)
Step-by-step Solution
Detailed explanation
Mass of \(C _{2} H _{5} OH =62.5 \times 0.8=50\,g\) \(\Delta T _{ f }= K _{ f } \times m\) \(0.9=2 \times \frac{1.8 \times 1000}{ M _{ \pi } \times 50}\) \(M _{ w }=\frac{2 \times 1.8 \times 1000}{0.9 \times 50}=80\)
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