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Likurg_2 [28]
2 years ago
11

Please solve ...................................................................

Chemistry
1 answer:
scZoUnD [109]2 years ago
7 0

Answer:

Explanation:

Part (c)

Mix together two suitable solutions.

Use filtration to separate the precipitate as a residue from the solution.

Wash the precipitate with distilled water while it is in the filter funnel.

Leave the washed precipitate aside or in a warm oven to dry.

Part (d)

Pb(NO3)2 (aq) + Na2SO4 (aq) → 2 NaNO3 (aq) + PbSO4 (s)

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Given that the freezing point depression constant for water is 1.86°c kg/mol, calculate the change in freezing point for a 0.907
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Answer : The correct answer for change in freezing point = 1.69 ° C

Freezing point depression :

It is defined as depression in freezing point of solvent when volatile or non volatile solute is added .

SO when any solute is added freezing point of solution is less than freezing point of pure solvent . This depression in freezing point is directly proportional to molal concentration of solute .

It can be expressed as :

ΔTf = Freezing point of pure solvent - freezing point of solution = i* kf * m

Where : ΔTf = change in freezing point (°C)

i = Von't Hoff factor

kf =molal freezing point depression constant of solvent.\frac{^0 C}{m}

m = molality of solute (m or \frac{mol}{Kg} )

Given : kf = 1.86 \frac{^0 C*Kg}{mol}

m = 0.907 \frac{mol}{Kg} )

Von't Hoff factor for non volatile solute is always = 1 .Since the sugar is non volatile solute , so i = 1

Plugging value in expression :

ΔTf = 1* 1.86 \frac{^0 C*Kg}{mol} * 0.907\frac{mol}{Kg} )

ΔTf = 1.69 ° C

Hence change in freezing point = 1.69 °C

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