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AP EAMCET · Chemistry · Solutions

What is the boiling point of solution of \(0.1 \mathrm{~m} \mathrm{KCl}\) ? \(\mathrm{K}_{\mathrm{b}}\) of water is \(0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \cdot(\alpha=100 \%)\)(water boil at \(373 \mathrm{~K}\) )

  1. A \(100.104 \mathrm{~K}\)
  2. B \(373.104 \mathrm{~K}\)
  3. C \(273.104 \mathrm{~K}\)
  4. D \(373.052 \mathrm{~K}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(373.104 \mathrm{~K}\)

Step-by-step Solution

Detailed explanation

\(\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \times \mathrm{m} \times \mathrm{i}=(0.52)(0.1)(2)=0.104 \mathrm{~K}\) \(\Rightarrow \mathrm{T}_{\mathrm{b}}=\mathrm{T}_{\mathrm{b}}^0+\Delta \mathrm{T}_{\mathrm{b}}=373+0.104=373.104 \mathrm{~K}\)
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