The molal elevation constant of water is ${0.51}^{\circ} C$ $K g m o l^{- 1}$ . The boiling point of 0.1 molal aqueous NaCl solution is nearly :

{100.05}^{\circ} C

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{100.1}^{\circ} C

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{100.2}^{\circ} C

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{101.0}^{\circ} C

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Solution

The correct option is B ${100.1}^{\circ} C$
Explanation :
Remember, each molecule of NaCl produces two ions, so the van't Hoff factor, i - is 2 here.

$Δ T_{b} = i K_{b} . m = 2 \times 0.51 \times 0.1 = {0.102}^{\circ}$

Now b.p. of solution = $100 + Δ T b$

= $100 + 0.102 = {100.102}^{\circ} C$

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The molal elevation constant of water = 0.52⁰C. The boiling point of 1.0 molal aqueous KCl solution (assuming complete dissociation of KCl), therefore, should be

Q. 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:

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The molal elevation constant of water is 0.51∘CKgmol−1 . The boiling point of 0.1 molal aqueous NaCl solution is nearly :

The correct option is B 100.1∘CExplanation : Remember, each molecule of NaCl produces two ions, so the van't Hoff factor, i - is 2 here. ΔTb=iKb.m=2×0.51×0.1=0.102∘ Now b.p. of solution = 100+ΔTb = 100+0.102=100.102∘C

The molal elevation constant of water is ${0.51}^{\circ} C$ $K g m o l^{- 1}$ . The boiling point of 0.1 molal aqueous NaCl solution is nearly :

The correct option is B ${100.1}^{\circ} C$
Explanation :
Remember, each molecule of NaCl produces two ions, so the van't Hoff factor, i - is 2 here.

$Δ T_{b} = i K_{b} . m = 2 \times 0.51 \times 0.1 = {0.102}^{\circ}$

Now b.p. of solution = $100 + Δ T b$

= $100 + 0.102 = {100.102}^{\circ} C$