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Paul [167]
3 years ago
10

What is the freezing point of an aqueous solution made by dissolving 0.118 mole of MgCl2 in 100.0 g of water? Kf = 1.86 oC/m

Chemistry
1 answer:
Sauron [17]3 years ago
8 0

Answer:

Freezing point of solution is -4.39°C

Explanation:

Freezing point depression's formula:

ΔT = Kf . m . i

ΔT = Freezing point of pure solvent - Freezing point of solution

0°C - X = 1.86°C/m . m . 2

i = Van't Hoff factor, ions dissolved in solution

MgCl₂ → Mg²⁺  + 2Cl⁻     i = 3  →  1 mol of Mg and 2 mol of chloride

Let's think the molality (mol/kg)

First of all, let's convert the mass of solvent from g to kg

100 g / 1000 = 0.1 kg

Molality (mol/kg) = 0.118 mol / 0.1 kg = 1.18 m

0°C - X = 1.86°C/m . 1.18 m . 2

X = - 4.39°C

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8 0
4 years ago
A 0.1 gram sample of an unknown liquid is vaporized completely at 70 degrees C to fill a 750mL flask. The pressure is 0.05951atm
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Answer:

The molar mass of the liquid 62.89 g/mol

Explanation:

Step 1: Data given

Mass of the sample = 0.1 grams

Temperature = 70°C

Volume = 750 mL

Pressure = 0.05951 atm

Step 2: Calculate the number of moles

p*V = n*R*T

n = (p*V)/(R*T)

⇒ with n = the number of moles gas = TO BE DETERMINED

⇒ with p = The pressure = 0.05951 atm

⇒ with V = The volume of the flask = 750 mL = 0.750 L

⇒ with R = The gasconstant = 0.08206 L*atm/K*mol

⇒with T = the temperature = 70 °C = 343 Kelvin

n = (0.05951 *0.750)/(0.08206*343)

n = 0.00159 moles

Step 3: Calculate molar mass

Molar mass = mass / moles

Molar mass =0.1 gram /  0.00159 moles

Molar mass = 62.89 g/mol

The molar mass of the liquid 62.89 g/mol

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