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Musya8 [376]
3 years ago
14

a solution containing 80 g of KCL in 200 g of H2O at 50 degrees celcius is cooled to 20 degrees celcius. How many grams of KCL r

emain in solution at 20 degrees celcius? How many grams of solid KCL crystallized after cooling?​
Chemistry
1 answer:
Varvara68 [4.7K]3 years ago
8 0

Answer:

12g KCl will be crystallized

Explanation:

To solve this problem you need to know solubility of KCl in water at 20°C is 34g per 100g of water.

That means the maximum concentration of KCl you can dissolve at 20°C in 200g of water is 34g×2 = 68g of KCl

As solution containing 80g of KCl, the extra KCl will be crystallized after cooling, that is:

80g of KCl - 68g of KCl = <em>12g KCl will be crystallized</em>

I hope it helps!

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Answer:

see explanation below

Explanation:

The question is incomplete. Here's the complete question:

<em>Consider the following reaction: </em>

<em>SO2Cl2 -----> SO2(g) + Cl2(g) </em>

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<em>If a reaction mixture initially contains 0.168 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?</em>

This is a problem of equilibrium, therefore, we need to solve this using the expression of equilibrium constant. To do that, we need to wirte an ICE chart and solve from there:

       SOCl2 ---------> SO2 + Cl2     Kc = 2.99x10⁻⁷

i)       0.168                  0        0

c)         -x                    +x       +x

e)     0.168-x                x         x

Writting the Kc expression:

Kc = [SO2] [Cl2] / [SOCl2]

Replacing the values from the chart:

2.99x10⁻⁷ = x² / 0.168 - x

However, Kc is a very very small value, therefore, we can assume that the value of "x" would be very small too, and we can neglect the 0.168-x and just round it to 0.168:

2.99x10⁻⁷ = x²/0.168

2.99x10⁻⁷ * 0.168 = x²

√5.02x10⁻⁸ = x

x = 2.24x10⁻⁴ M

This means then, that the concentration of Cl2 in equilibrium would be:

<em>[Cl₂] = 2.24x10⁻⁴ M</em>

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