Based on the effect of Raoult\'s Law, what liquid will boil first? A or B Explai

Question

Based on the effect of Raoult's Law, what liquid will boil first? A or B Explain: To get the solution in beaker B to boil you will have to (increase/decrease the temperature? K_b = boiling point elevation constant Each solvent has its own value of K_b m_solution = molality of the solution Based on that effect we can say that: DeltaT_b = K_b x m_solution. Is the boiling point of a 0.0500 m Kl aqueous solution higher, lower or the same as a 0.0500 m glucose solution? Explain (No calculations needed)

Explanation / Answer

The boiling point of a pure solvent is lower than the boiling point of its solution containing nonvolatile solute

The presence of non volatile solute (glucose in this case) decreases the vapour pressure as less number of solvent molecules get surface to escape out of liquid surface.

The boiling point is the temperature where the vapour pressure of liquid becomes equal to atmospheric pressure.

so due to less vapour pressure of solvent with solute dissolved in it, we need to heat it up at higher temperature so that more number of molecules may escape from the surface of liquid.

Hence the boiling point increases.

It is expressed as given in the equation

Delta Tb = Kb X molality of solution

The molality = moles of solute / mass of solvent in Kg

So more the number of solutes more the molality and hence more the elevation in boiling point.

Now in case of KI it will dissociates as K+ and I- so each mole of KI will give two ions while in case of glucose (non ionizable) the number of molecules will be the same as dissolved

So we have to multiply the molality of KI by two to get the exact molality of solution

Which increases the elevation in boiling point twice that of glucose solution.

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