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kupik [55]
3 years ago
9

. The freezing point of an aqueous solution containing a nonelectrolyte solute is – 2.79 °C. What is the boiling point of this s

olution?
Chemistry
1 answer:
Semmy [17]3 years ago
4 0

Answer:

Boiling point of the solution is 100.78°C

Explanation:

This is about colligative properties.

First of all, we need to calculate molality from the freezing point depression.

ΔT = Kf . m . i

As the solute is nonelectrolyte, i = 1

0°C - (-2.79°C) = 1.86 °C/m . m . 1

2.79°C / 1.86 m/°C = 1.5 m

Now, we go to the boiling point elevation

ΔT = Kb . m . i

Final T° - 100°C  =  0.52 °C/m . 1.5m . 1

Final T° =  0.52 °C/m . 1.5m . 1  + 100°C → 100.78°C

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Answer:

Number of mole = 13.155moles

Explanation:

One mole of a substance is equal to 6.022 * 10^23 units of that substance.

Mole of a substance can be defined according to chemistry as the mass of substance containing the same number of fundamental units

Using the formula for mole

n = m/Mm

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m = number of mass

Mm = number of molar mass

We are provided with some information

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Molar mass of C = 12.0107g/mol

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Calculate the initial rate for the formation of c at 25 ∘c, if [a]=0.50m and [b]=0.075m
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If two products, call them A and B react together to form product C, a general equation for the formation of C has the form:

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Normally you run controled trials in lab which permit to calculate k, m and n .

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<span>2         0.50    0.020       6.0×10−3 </span>
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Given that for trials 1 and 2 [A] is the same you can use those values to find n, in this way

rate 1 = 3.0 * 10^ -3 = k [A1]^m * [B1]^n

rate 2 = 6.0*10^-3 = k [A2]^m * [B2]^n

divide rate / rate 1 => 2 = [B1]^n / [B2]^n

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Given that for data 1 and 3 [B] is the same, you use those data to find m

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4 = 2^m => m = 2

Now use any of the data to find k

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rate = k[A]^2*[B] = 0.12 m^-2 s^-1 * (0.50m)^2 * (0.075m) = 0.0045 m/s = 4.5*10^=3 m/s

Answer: 4.5 * 10^-3 m/s
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