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Masja [62]
3 years ago
13

The boiling point elevation of an aqueous sucrose is found to be 0.39 Celsius. What mass of sucrose (molar mass = 342.30g/mol) w

ould be needed to dissolve in 500.0 g of water? Kb (water) = 0.513 C/m
Chemistry
1 answer:
FromTheMoon [43]3 years ago
7 0

Answer:

130 g of sucrose

Explanation:

Boiling point elevation formula → ΔT = Kb . m

ΔT = Boiling T° solution - Boiling T° pure solvent → 0.39°C

0.39°C = 0.513°C/m . M

m = 0.760 mol/kg → molality = moles of solute / 1kg of solvent

Let's determine the moles of solute → molality . kg

0.760 mol/kg. 0.5 kg = 0.380 moles

If we convert the moles to mass, we'll get the answer

0.380 mol . 342.30 g/mol = 130g

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Answer:

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Explanation:

The equation for the equilibrium is

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Now, we can set up an ICE table

                     CN^- + H2O ⇌ HCN + OH^-

I/(mol/L)      0.255                     0         0

C/(mol/L)       -x                        +x        +x

E/(mol/L)  0.255 - x                   x         x

Ka = x^2/(0.255 - x) = 2.05 × 10^-5

Check for negligibility

0.255/(2.05 × 10^-5) = 12 000 > 400. ∴ x ≪ 0.255

    x^2 = 0.255(2.05 × 10^-5) = 5.20 × 10^-6

        x = sqrt(5.20 × 10^-6)    = 2.28 × 10^-3

[OH^-] = x mol/L                     = 2.28 × 10^-3 mol/L

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