Question

How much would the boiling point of water increase if 4 mol of NaCl were

Answer 1

Answer:

 ΔT = 8.51°C

Explanation:

Data Given:

Amount of NaCl = 4 mol

Amount of water = 1 kg

Kb = 0.51°C/(mol/kg)

Increase in boiling point = ?

In this case elevation of boiling point will be determined

For this purpose the Formula will be Used

                                   ΔT =Kb.m(solute) -----------------------(a)

Where

ΔT = boiling point elivation (°C)

Kb = molal boiling point elevation constant (°C/(mol/kg))

m (solute) = molality of the solute (mole of solute/ kg solvent)

we have to first find the Molality of NaCl in order to calculate ΔT

Formula for Molality

                         Molality = mole of NaCl / kg of water ------------------ (1)

Put Values in equation 1

                       Molality = 4 mol/ 1 kg

                       Molality = 4 m

Note ** m is for molality

NaCl dissociates into it ions Na⁺ and Cl⁻

The ionization factor of Na⁺  (i) = 2

* That is 2 mole of solute dissoled pre mole

for Na⁺ molality total = m x i

                          m (total) = 4 x 2 = 8 m ---------------------(3)

Put the vlaue from given date and m from equation (3) in equation (a)

                                 ΔT =Kb.m(solute)

                                 ΔT = 0.51°C/(mol/kg) x 8 m

                                ΔT = 8.51°C

Where the unit of molality (m) is mol/kg that is cancel out in equation

So, the Increase in Temperature is 8.51°C