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deff fn [24]
3 years ago
12

500 mL of water is added to 400 mL of 0.35 M HCl. Find the concentration of the diluted solution.

Chemistry
1 answer:
jarptica [38.1K]3 years ago
5 0

The concentration of diluted solution is 0.16 M

<u>Explanation:</u>

As, the number of moles of diluted solution and concentrated solution will be same.

So, the equation used to calculate concentration will be:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the concentrated HCl solution

M_2\text{ and }V_2 are the molarity and volume of diluted HCl solution

We are given:

M_1=0.35M\\V_1=400mL\\M_2=?M\\V_2=(500+400)mL=900mL

Putting values in above equation, we get:

0.35\times 400=M_2\times 900\\\\M_2=\frac{0.35\times 400}{900}=0.16M

Hence, the concentration of diluted solution is 0.16 M

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Answer:

The answer to your question is d. 0.5 M

Explanation:

Data

[A] = 1M

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Concentration of B and C at equilibrium = x

Concentration of A at equilibrium = 1 - x

Equation of equilibrium

                                       k = \frac{[B][C]}{A}

Substitution

                                       0.5 = \frac{[x][x]}{1 - x}

Simplification

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Solve for x

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                                      0.5 - 0.5x = x²

                                       x² + 0.5x - 0.5 = 0

Find the roots             x₁ = 0.5     x₂ = -1

There are no negative concentrations so the concentration of A at equilibrium is  

                       [A] = 1 - 0.5

                             = 0.5 M

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