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Elevation in Boiling Point MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • Elevation in Boiling Point is considered one the most difficult concept.

  • 22 Questions around this concept.

Solve by difficulty

At 100°C the vapour pressure of a solution of 6.5g of a solute in 100g water is 732 mm. If Kb=0.52, the boiling point of this solution will be:

Assertion: The boiling point of a solution increases with increasing concentration of the solute.

Reasoning: The addition of a solute to a solvent increases the boiling point of the resulting solution due to the colligative property of boiling point elevation.

 

 

Assertion: A solution with a higher molality will always exhibit a larger boiling point elevation than a solution with a lower molality.

Reasoning: The boiling point elevation of a solution is directly proportional to the molality of the solution.

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A solution is prepared by dissolving a non-volatile solute in a solvent. Which of the following will result in a greater elevation in boiling point of the solution?

The rinse in boiling point of solution containing 1.4 g glucose in 100 g of solvent in 0.8^{\circ} \mathrm{C}. The molal elevation constant of liquid

A solution containing  4 \mathrm{~g}  of non volatile solute What will be the \mathrm{m}  olecular mass of the solute if solute is in  \mathrm{30 \mathrm{~g}} of water boils at  \mathrm{373.52 \mathrm{k}} 
{ Water boils at \mathrm{ 373 \mathrm{k}, \mathrm{Kb}}   of water  \mathrm{ =0.52 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}} }

Calculate the molar mass of substance \mathrm{" X"}  when 5gYX is dissolved in  \mathrm{100 \mathrm{~g}}  of  \mathrm{\mathrm{CCl}_4} , it raised boiling point by \mathrm{0.90 \mathrm{~K}. \{ \mathrm{Kb}}  for \mathrm{ \mathrm{CCl}_4}  is  5.0 \mathrm{~K} \mathrm{~{kgmol}^{-1}}}

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Concepts Covered - 1

Elevation in Boiling Point

Boiling Point: It is the temperature of a liquid at which its vapour pressure becomes equal to the atmospheric pressure. As vapour pressure of solution is lower than that of the solvents. As on addition of a non-volatile solute in a pure solvent's V.P decreases or V.P of solvent in pure state is more than that of solution. Now in order to boil the solution it is necessary to increase the temperature of solution above the B.P of pure solvent. It means B.P. of solution is always higher than B.P. of pure solvent. This increase in B.P. of solution is called elevation in B.P.(ΔTb).

  • It is also termed as Ebullioscopy.
  • It is measured by Lands Berger's method and Koltrell's method.
  • Suppose Toand Tare the B.P. of pure solvent and solution respectively than elevation in B.P (ΔTb) is given as ΔT= Tb - Tob
    Here ΔTis directly propotional to molality of solution
    \begin{array}{l}{\Delta \mathrm{T}_{\mathrm{b}} \propto \mathrm{m}} \\ {\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \mathrm{m}}\end{array}
    If molality of the solution is one, then
    \Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}}
    The elevation in B.P. is also given as
    \Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \times \frac{\mathrm{W}}{\mathrm{M}} \times \frac{1000}{\mathrm{W}}
    Molecular weight of solute can be find out as follows
    \mathrm{M=\frac{K_{b} \times w \times 1000}{\Delta T_{b} \times W}}
    Here w = weight of solute
    W = weight of solvent
    K= molal elevation constant or ebullioscopic constant.
    M = molar mass of solute
    \mathrm{K}_{\mathrm{b}}=\frac{\mathrm{R} \mathrm{T}^{2}}{1000 \: \mathrm{Lv} \: \mathrm{or}\: \Delta \mathrm{Hv}}
    Here Lv or ΔHv = latent heat of vapourization.

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Elevation in Boiling Point

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