Question

2g. each of 2 solutes A and B (molecular mass of A>B) are dissolved separately in 50g.each of the same solvent. Which will show greater elevation in Boiling point?

Answer 1

elevation of boiling point is inversely proportional to molecular mass

it is directly proportional to molality but as molality has mass in division the above case occurs

as mass of a greater that b it has lower elevation in boiling point


please give brainliest

Answer 2

Answer: solute B

Explanation:

Weight of solvent = 50 g = 0.05 kg (1 kg=1000 g)

Mass of solute added = 2 g

1. For solute A:

$\Delta T_b=K_b\times \frac{\text{mass of solute A}}{\text{molar mass of solute A}\times \text {weight of solvent in kg}}$

$\Delta T_b=K_b\times \frac{2g}{m_A\times 0.05kg}$     (1)

$\Delta T_b^A$ = elevation in boiling point

$K_b$ = boiling point constant

2.  For solute B:

$\Delta T_b=K_b\times \frac{\text{mass of solute B}}{\text{molar mass of solute B}\times \text{weight of solvent in kg}}$

$\Delta T_b^B=K_b\times \frac{2g}{m_B\times 0.05kg}$ (2)

On dividing 2 by 1

$\frac{\Delta T_b^B}{\Delta T_b^A}=\frac{M_A}{M_B}$

Now as given $M_A$ >$M_B$

$\Delta T_b^A}$ < $\Delta T_b^B$