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Lilit [14]
3 years ago
14

What is the molarity of a stock solution if 60 mL were used to make 150 mL of a .5M solution

Chemistry
1 answer:
kirza4 [7]3 years ago
6 0

Answer:

The answer to your question is 1.25 M

Explanation:

Data

Molarity 1 = ?

Volume 1 = 60 ml

Molarity 2 = 0.5 M

Volume 2 = 150 ml

Process

1.- Write the dilution formula

       Molarity 1 x Volume 1 = Molarity 2 x Volume 2

-Solve for Molarity 1

       Molarity 1 = Molarity 2 x Volume 2 / Volume 1

-Substitution

        Molarity 1 = (0.5)(150) / 60

-Simplification

        Molarity = 75 / 60

-Result

        Molarity = 1.25 M

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andreyandreev [35.5K]

<u>Answer:</u> The correct answer is Option 5.

<u>Explanation:</u>

  • To calculate the molarity of the solution after mixing 2 solutions, we use the equation:

M=\frac{n_1M_1V_1+n_2M_2V_2}{V_1+V_2}

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of the NaOH.

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of the Ca(OH)_2

We are given:

n_1=1\\M_1=0.10M\\V_1=50mL\\n_2=2\\M_2=0.1\\V_2=50mL  

Putting all the values in above equation, we get:

M=\frac{(1\times 0.1\times 50)+(2\times 0.1\times 50)}{50+50}\\\\M=0.15M

  • To calculate the molarity of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HCl

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base.

We are given:

n_1=1\\M_1=?M\\V_1=250mL\\n_2=1\\M_2=0.15M\\V_2=100mL

Putting values in above equation, we get:

1\times M_1\times 250=1\times 0.15\times 100\\\\M_1=0.06M

Hence, the correct answer is Option 5.

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