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Sever21 [200]
3 years ago
10

A solution of LiCl in water has XLiCl = 0.0800. What is the molality? A solution of LiCl in water has XLiCl = 0.0800. What is th

e molality? 4.44 m LiCl 8.70 m LiCl 4.83 m LiCl 4.01 m LiCl
Chemistry
1 answer:
Kay [80]3 years ago
7 0

Answer:

mol LiCl = 4.83 m

Explanation:

GIven:

Solution of LiCl in water XLiCl = 0.0800

Mol of water in kg = 55.55 mole

Find:

Molality

Computation:

mole fraction = mol LiCl / (mol water + mol LiCl)

0.0800 = mol LiCl / (55.55 mol + mol LiCl)

0.0800 mol LiCl + 4.444 mol = mol LiCl

mol LiCl - 0.0800 mol LiCl = 4.444 mol

0.92 mol LiCl = 4.444 mol

mol LiCl = 4.83 m

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The equilibrium constant for the reaction
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The question is incomplete, here is the complete question:

The equilibrium constant for the reaction

N₂O₄(g)⇌2NO₂ at 2°C is Kc = 2.0

If each yellow sphere represents 1 mol of N₂O₄(g) and each gray sphere 1 mol of NO₂ which of the following 1.0 L containers represents the equilibrium mixture at 2°C?

The image is attached below.

<u>Answer:</u> The system which represents the equilibrium having value of K_c=2.0 is system (b)

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_c

For a general chemical reaction:

aA+bB\rightarrow cC+dD

The expression for K_{c} is written as:

K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}

For the given chemical equation:

N_2O_4(g)\rightleftharpoons 2NO_2

The expression of K_c for above equation follows:

K_c=\frac{[NO_2]^2}{[N_2O_4]}      .......(1)

We are given:

Volume of the container = 1.0 L

Value of K_c = 2.0

Molarity of the substance is calculated by using the equation:

\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume}}

For the given images:

  • <u>For a:</u>

Number of Gray spheres = 8 moles

Number of yellow spheres = 4 moles

Putting values in expression 1, we get:

K_c=\frac{(8/1)^2}{(4/1)}\\\\K_c=16

  • <u>For b:</u>

Number of Gray spheres = 4 moles

Number of yellow spheres = 8 moles

Putting values in expression 1, we get:

K_c=\frac{(4/1)^2}{(8/1)}\\\\K_c=2

  • <u>For c:</u>

Number of Gray spheres = 6 moles

Number of yellow spheres = 6 moles

Putting values in expression 1, we get:

K_c=\frac{(6/1)^2}{(6/1)}\\\\K_c=6

  • <u>For d:</u>

Number of Gray spheres = 2 moles

Number of yellow spheres = 8 moles

Putting values in expression 1, we get:

K_c=\frac{(2/1)^2}{(8/1)}\\\\K_c=\frac{1}{2}

Hence, the system which represents the equilibrium having value of K_c=2.0 is system (b)

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