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mel-nik [20]
3 years ago
14

g Sucrose (C12H22O11), a nonionic solute, dissolves in water (normal freezing/melting point 0.0°C) to form a solution. If some u

nknown mass of sucrose is dissolved in 150.g of water and this solution has a freezing point of -0.56°C, calculate the mass of sucrose dissolved.
Chemistry
1 answer:
DIA [1.3K]3 years ago
4 0

Answer:

15.4 g of sucrose

Explanation:

Formula to be applied for solving these question: colligative property of freezing point depression. → ΔT = Kf . m

ΔT = Freezing T° of pure solvent - Freezing T° of solution

Let's replace data given: 0°C - (-0.56°C) = 1.86 C/m . m

0.56°C / 1.86 m/°C = m → 0.301 mol/kg

m → molality (moles of solute in 1kg of solvent)

Our mass of solvent is not 1kg, it is 150 g. Let's convert it from g to kg, to determine the moles of solute: 150 g. 1kg/1000g = 0.150 kg

0.301 mol/kg .  0.150kg = 0.045 moles.

We determine the mass of sucrose, by the molar mass:

0.045 mol . 342 g/1mol = 15.4 g

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kolbaska11 [484]

Answer:

469.9K

Explanation:

Using Charles's law equation:

V1/T1 = V2/T2

Where;

V1 = initial volume (ml)

V2 = final volume (ml)

T1 = initial temperature (K)

T2 = final temperature (K)

Based on the information provided:

V1 = 280.0ml

V2 = 440.0 ml

T1 = 26.0°C = 26 + 273 = 299K

T2 = ?

Using V1/T1 = V2/T2

280/299 = 440/T2

0.936 = 440/T2

T2 = 440 ÷ 0.936

T2 = 469.9K

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M = 4.0 moles/2 Liters

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