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expeople1 [14]
3 years ago
15

A solution of H2SO4 (90.08 g/mol) has 785.6 g of H2SO4 and 359 g of H2O. What is the molal concentration (molality)?

Chemistry
1 answer:
tekilochka [14]3 years ago
8 0

The molality of H₂SO₄ solution is 24.2 m.

<u>Explanation:</u>

We need to find the molality of sulfuric acid.

Mass of sulfuric acid = 785.6 g

Mass of water = 359 g

We have to find the moles of H₂SO₄ by using its mass and molar mass as,

Moles of H₂SO₄ = $\frac{785.6g}{90.08 g/mol}

                         = 8.7 moles

Mass of the solution in kg =$\frac{359}{1000} =  0.359 kg

Molality = $\frac{moles}{solvent (kg)}

           = $\frac{8.7 mol }{0.359 kg}\\

          = 24.2 m

So molality of the H₂SO₄ solution is 24.2 m.

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Consider the following system at equilibrium where AH° = 10.4 kJ, and K. = 1.80x102 at 698 K. 2HI(2) P H2() +132) If the VOLUME
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<u>Answer:</u>

<u>For 1:</u> The correct answer is Option C.

<u>For 2:</u> The correct answer is Option B.

<u>For 3:</u> The correct answer is Option C.

<u>For 4:</u> The correct answer is Option C.

<u>Explanation:</u>

We are given:

K_e=1.80\times 10^2

\Delta H^o_{rxn}=10.4kJ

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Any change in the equilibrium is studied on the basis of Le-Chatelier's principle. This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

Any change in volume during equilibrium is studied on the basis of change of moles of reactants and products.

When volume is decreased, the equilibrium will shift in the direction which produces fewer moles of gas and when volume is increased, the equilibrium will shift in the direction which produces more moles of gas.

Number of moles of gases on reactant side = [1 + 1] = 2

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Change in number of moles = 2 -2 = 0

As, number of moles on both the side of the reaction is same. This means there will be no effect on change in volume.

  • <u>For 1:</u>

As, there is no effect of volume on the equilibrium. So, the equilibrium constant remains the same.

Hence, the correct answer is Option C.

  • <u>For 2:</u>

K_c is the constant of a certain reaction at equilibrium while Q_c is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

There are 3 conditions:

  • When K_{e}>Q_c; the reaction is product favored.
  • When K_{e}; the reaction is reactant favored.
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As, the value of K_e is not changing.

This means that value of Q_c will be equal to K_e

Hence, the correct answer is Option B.

  • <u>For 3:</u>

As, the equilibrium constant is not changing. So, the reaction is present at equilibrium.

Hence, the correct answer is Option C.

  • <u>For 4:</u>

The reaction is present at equilibrium. So, the concentration of iodine gas will remain the same.

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