If 1 mole of NaCl solute is dissolved into the 1 kg of water, at what temperature will water boil at 1.013 bar? (Kb of water is 0.52 K kg mol-1)

Question

If 1 mole of NaCl solute is dissolved into the 1 kg of water, at what temperature will water boil at 1.013 bar? (Kb of water is 0.52 K kg mol-1) (A) 373.15 K (B) 373.67 K (C) 374.19 K (D) 373.19 K

Tardigrade Admin · Accepted Answer

∵ Δ T= Kb (n/wA) Given, Kb =0.52 K kg mol -1 n =1, wA=1 kg ∴ Δ T =(0.52 × 1/1)=0.52 K ∵ Boiling point of pure water at 1.013 bar (i.e. 1 atm ) is 373.15 K ∴ Boiling point of the solulion =373.15+0.52 =373.67 K

Q. If $1 m o l e$ of NaCl solute is dissolved into the $1 k g$ of water, at what temperature will water boil at 1.013 bar? ( $K_{b}$ of water is $0.52 K k g m o l^{- 1}$ )

373.15 K

373.67 K

374.19 K

373.19 K

375 K

$∵ Δ T = K_{b} \frac{n}{w _{A}}$

Given, $K_{b} = 0.52 K k g m o l^{- 1}$

$n = 1, w_{A} = 1 k g$

$∴ Δ T = \frac{0.52 \times 1}{1} = 0.52 K$

$∵$ Boiling point of pure water at $1.013$ bar (i.e. $1 a t m$ ) is $373.15 K$

$∴$ Boiling point of the solulion $= 373.15 + 0.52$

$= 373.67 K$

Q. If 1mole of NaCl solute is dissolved into the 1kg of water, at what temperature will water boil at 1.013 bar? (Kb​ of water is 0.52Kkgmol−1)