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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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As per the question we have to understand the role of strong nuclear force and weak nuclear force.

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Present value lets us compare dollar values from different time periods.
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3 years ago
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Romashka-Z-Leto [24]

Answer:

45.3°C

Explanation:

Step 1:

Data obtained from the question.

Initial pressure (P1) = 82KPa

Initial temperature (T1) = 26°C

Final pressure (P2) = 87.3KPa.

Final temperature (T2) =.?

Step 2:

Conversion of celsius temperature to Kelvin temperature.

This is illustrated below:

T(K) = T(°C) + 273

Initial temperature (T1) = 26°C

Initial temperature (T1) = 26°C + 273 = 299K.

Step 3:

Determination of the new temperature of the gas. This can be obtained as follow:

P1/T1 = P2/T2

82/299 = 87.3/T2

Cross multiply to express in linear form

82 x T2 = 299 x 87.3

Divide both side by 82

T2 = (299 x 87.3) /82

T2 = 318.3K

Step 4:

Conversion of 318.3K to celsius temperature. This is illustrated below:

T(°C) = T(K) – 273

T(K) = 318.3K

T(°C) = 318.3 – 273

T(°C) = 45.3°C.

Therefore, the new temperature of the gas in th tire is 45.3°C

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