Question

Can someone please help me please?

Answer 1

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

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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

$\Delta T_{f} = 1.86 \times 0.1404$

$\Delta T_{f} = 0.261 \textdegree \text{C}$

Answer 2

Answer:

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

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

Explanation:

1) $\Delta T_b=T_b-T$

$\Delta T_b=K_b\times m$

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

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

$K_b=?$

$\Delta T_b=K_b\times m$

$2.4^oC=K_b\times 3.1 m$

$K_b=\frac{2.4 ^oC}{3.1 m}=0.07742 ^oC/m$

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

2) $\Delta T_f=T-T_f$

$\Delta T_f=K_f\times m$

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

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 = $\frac{0.705 mol}{5.02 kg}=0.1404 mol/kg$

$K_f=1.86^oC/m$

$\Delta T_f=iK_f\times m$

$=1\times 1.86 ^oC/m\times 0.1404 m$

$=0.261^oC$

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