If the solution boils at a temperature $T_{1}$ and the solvent at a temperature $T_{2}$ , the elevation of boiling point is:

T_{1}

T_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

T_{1}

T_{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

T_{2}

T_{1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

T_{1}

\times

T_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $T_{1}$ - $T_{2}$
The elevation in the boiling point will be $T_{1} - T_{2}$ .
The reason is after adding non-volatile solute the boiling point of the solution will increase.
Vapour pressure of the solution at some temperature is found to be lower than the vapour pressure of the pure solvent at the same temperature. In such a solution, the nonvolatile solute particles will obstruct the escaping of solvent molecules from the surface.
Therefore, it reduces the vapour pressure. Due to this, Elevation in the boiling point is based on the colligative property.
Hence, option B is correct.

Suggest Corrections

Similar questions

If the solution boils at a temperature T₁ and the solvent at a temperature T₂the elevation of boiling point is given by

I, I I, I I I

T_{1}, T_{2}

T_{3}

T_{1}, T_{2}

T_{3}

I, II and III are three isotherms respectively at $T_{1}$ , $T_{2}$ and $T_{3}$ temperatures. The order of temperature will be

Join BYJU'S Learning Program