One of the reason salt is added to water is because it increases the boiling point of the water, so it will cook better.

How much salt NaCl(s) would have to be dissolved in 1.00 L of water in order to raise the boiling point by 1°C?

Data: boiling point elevation constant of water = 0.513 K.kg.mol^-1

Question

One of the reason salt is added to water is because it increases the boiling point of the water, so it will cook better.

How much salt NaCl(s) would have to be dissolved in 1.00 L of water in order to raise the boiling point by 1°C?

Data: boiling point elevation constant of water = 0.513 K.kg.mol^-1

| General Chemistry 2

Accepted Answer

The magnitude of the boiling point elevation ΔT_b (in K) is:

ΔT_b = i x K_b x m

ΔT_b = T_{b NaCl solution} – T_{b water}

i = van’t Hoff i-factor = number of ions produced per formula unit

m = solution molality (in mol.kg^-1)

K_b = boiling point elevation constant of water = 0.513 K.kg.mol^-1

We want ΔT_b = 1°C = 1 K

NaCl (s) + H₂O (l) → Na⁺ (aq) + Cl^- (aq)

⇒ i = 2

ΔT_b = i x K_b x m

⇒ m = $\frac{∆ T_{b}}{i \times K_{b}}$ = 0.975 mol.kg^-1

m =

\frac{n_{NaCl}}{m_{water}}

=

\frac{m_{NaCl}}{M_{NaCl} \times m_{water}}

m_NaCl= m x M_NaClx m_water = 0.975 x 58.44 x 1.00 = 57.0 g

Exercise 7 | Colligative Properties of Solutions