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lora16 [44]
2 years ago
13

Calculate the freezing point (0°C) of a 0.05500 m aqueous solution of glucose.

Chemistry
1 answer:
Rainbow [258]2 years ago
4 0

Answer:

-0.1767°C (Option A)

Explanation:

Let's apply the colligative property of freezing point depression.

ΔT = Kf . m. i

i = Van't Hoff factot (number of ions dissolved). Glucose is non electrolytic so i = 1

m = molality (mol of solute / 1kg of solvent)

We have this data → 0.095 m

Kf is the freezing-point-depression constantm 1.86 °C/m, for water

ΔT = T° frezzing pure solvent - T° freezing solution

(0° - T° freezing solution) = 1.86 °C/m . 0.095 m . 1

T° freezing solution = - 1.86 °C/m . 0.095 m . 1 → -0.1767°C

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julsineya [31]

The reaction between butanoic acid and ethanol is:

CH3CH2CH2COOH + CH3CH2OH → CH3CH2CH2COOCH3CH2 + H2O

Based on the reaction stoichiometry:

1 mole of butanoic acid forms 1 mole of ethyl butyrate

Now,

Molar mass of Butanoic acid = 88.0 g/mol

Given mass of butanoic acid = 7.20 g

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# moles of ethyl butyrate formed = 0.0818 moles

Molar mass of ethyl butyrate = 116 g/mol

Mass of ethyl butyrate synthesized = 0.0818 * 116 = 9.49 g



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Explanation:

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<em></em>

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