Question

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13. The boiling point of a solvent is elevated by 2.4 °C when the solute concentration is 3.1 m. What is Kb?


What is the freezing-point depression of a solution that contains 0.705 mol of a nonelectrolyte solute in 5.02 kg of water? (Kf = 1.86 °C/m)

Answer 1

1) The value of the
`K_b` is 0.07742°C/m.

2) 0.261°C is the freezing-point depression of a solution.

Step-by-step explanation:

1)
`\Delta T_b=T_b-T`

`\Delta T_b=K_b* m`

`\Delta T_b=iK_b* \frac{\text{Mass of solute}}{\text{Molar mass of solute}* \text{Mass of solvent in Kg}}`

where,

`\Delta T_b` =Elevation in boiling point

`K_b` = boiling point constant of solvent= 3.63 °C/m

1 - van't Hoff factor (non-electrolyte solute)

m = molality

We have :
`\Delta T_b=2.4^oC`

m = 3.1 m

`K_b=?`

`\Delta T_b=K_b* m`

`2.4^oC=K_b* 3.1 m`

`K_b=(2.4 ^oC)/(3.1 m)=0.07742 ^oC/m`

The value of the
`K_b` is 0.07742°C/m.

2)
`\Delta T_f=T-T_f`

`\Delta T_f=K_f* m`

`Delta T_f=iK_f* \frac{\text{Mass of solute}}{\text{Molar mass of solute}* \text{Mass of solvent in Kg}}`

where,

`\Delta T_f` =depression in freezing point

`K_f` = freezing point constant of solvent= 1.86°C/m

1 - van't Hoff factor (non-electrolyte solute)

m = molality

We have , Moles of solute = 0.705 mol

Mass of solvent = 5.02 kg

`molality=\frac{\text{Moles of solute }}{\text{Mas of solvent(kg)}}`

m =
`(0.705 mol)/(5.02 kg)=0.1404 mol/kg`

`K_f=1.86^oC/m`

`\Delta T_f=iK_f* m`

`=1* 1.86 ^oC/m* 0.1404 m`

`=0.261^oC`

0.261°C is the freezing-point depression of a solution.

Answer 2

Answers:

0.77 °C·kg·mol⁻¹; 0.261 °C

Step-by-step explanation:

13. a. Boiling point elevation

The formula for boiling point elevation ΔTb is

ΔTb = Kb·b Divide each side by b

Kb = ΔTb/b

Kb = 2.4/3.1

Kb = 0.77 °C·kg·mol⁻¹

===============

13. b. Freezing point depression

The formula for freezing point depression
`\Delta T_(f)` is

`\Delta T_(f) = \Delta K_(f) \cdot b`

b = moles of solute/kilograms of solvent

b = 0.705/5.02

b = 0.1404 mol/kg

`\Delta T_(f) = 1.86 * 0.1404`

`\Delta T_(f) = 0.261 \textdegree \text{C}`