The molal elevation constant of water $= 0.52 K m^{- 1}$ . The boiling point of 1.0 molal aqueous KCl solution (assuming complete dissociation of $K C l$ ) should be:

{100.52}^{o} C

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{101.04}^{o} C

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{99.48}^{o} C

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{98.96}^{o} C

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Solution

The correct option is C ${101.04}^{o} C$
Given,

$k_{b} =$ molal elevation constantof water $= 0.52 K m o l^{- 1}$
$m =$ mollity= 1.0 m
$i =$ van't Hoff factor for KCl = 2

$Δ T_{b} = K_{b} \times m \times i$

$= 0.52 \times 1 \times 2 = 1.04$ ;

$∴ T_{b} = 100 + 1.04 = {101.04}^{o} C$

Hence, the correct option is $B$

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The molal elevation constant of water =0.52 K m−1. The boiling point of 1.0 molal aqueous KCl solution (assuming complete dissociation of KCl) should be:

The correct option is C 101.04oCGiven,kb= molal elevation constantof water=0.52 K mol−1m=mollity= 1.0 mi= van't Hoff factor for KCl = 2ΔTb=Kb×m×i=0.52×1×2=1.04;∴Tb=100+1.04=101.04oCHence, the correct option is B

The molal elevation constant of water $= 0.52 K m^{- 1}$ . The boiling point of 1.0 molal aqueous KCl solution (assuming complete dissociation of $K C l$ ) should be: