Verified Solution • 2025–26 Session moderate

QJEEAdv21Chb15c: JEE Advanced 2021 Chemistry - Class 12 JEEAdv21Chb15c Solution: The value of x is ________.

The value of x is ________.

Step-by-Step Verified Solution

amp;AgNO3(aq)⟶Ag+(aq)0.1 m+NO3−(aq)0.1 mamp;ΔTb=0.2×0.5amp;=0.1∘C=0.1 K\begin{aligned}& \mathrm{AgNO}_3(\mathrm{aq}) \longrightarrow \underset{0.1 ~ m}{\mathrm{Ag}^{+}(\mathrm{aq})}+\underset{0.1 ~m}{\mathrm{NO}_3^{-}(\mathrm{aq})}\\\\& \Delta \mathrm{T}_{\mathrm{b}}=0.2 \times 0.5 \\\\& =0.1^{\circ} \mathrm{C}=0.1 \mathrm{~K} \end{aligned}

Boiling point of solution =100.1∘C=X=100.1^{\circ} \mathrm{C} =\mathrm{X}

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