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Anastasy [175]
3 years ago
14

Can someone please help me please?

Chemistry
2 answers:
shepuryov [24]3 years ago
7 0

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 \times 0.1404

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

artcher [175]3 years ago
7 0

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.

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