Question

LUUS

What would the final boiling point of water be if 3 mol of NaCl were added to
1 kg of water (Kb = 0.51°C/(mol/kg) for water and i = 2 for NaCl)?

Answer 1

Final boiling point of water is 376·06 K

Step-by-step explanation:

As NaCl is a non-volatile solute so by the addition of non-volatile solute to a pure solvent will increase its boiling point

Formula for the change in boiling point of the solution is

ΔT
`x_(b)` = i × K
`x_(b)` × m

where

ΔT
`x_(b)` is the change in boiling point of the solution

i is the vant Hoff factor which is the number of particles per single molecule of the solute

K
`x_(b)` is the cryoscopic constant which only depends on the properties of the solvent and not on the solute

m is the molality of the solute

in this case molality of the NaCl is 3÷1 =3 mole/kg

By the formula

ΔT
`x_(b)` = 2×0·51×3=3·06 K

Boiling point of the pure solvent( in this case it is water) is 373 K

Final boiling point of the water = ΔT
`x_(b)` + 373

= 3·06+373

∴ Final boiling point of the water = 376·06 K