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Anna007 [38]
3 years ago
8

Se disuelven 10 g de acido carbónico H2CO3 en 150 g de agua. Determina la concentración % m/m de esa disolución?

Chemistry
1 answer:
sergey [27]3 years ago
6 0

The question is: 10 g of carbonic acid H2CO3 are dissolved in 150 g of water. Determine the% m / m concentration of that solution?

Answer: The% m / m concentration of that solution is 6.66%.

Explanation:

Given: Mass of solute = 10 g

Mass of solvent = 150 g

Formula used to calculate the %m/m is as follows.

Percent (m/m) = \frac{mass of solute}{mass of solvent} \times 100

Substitute the values into above formula as follows.

Percent (m/m) = \frac{mass of solute}{mass of solvent} \times 100\\= \frac{10 g}{150 g} \times 100\\= 6.66%

Thus, we can conclude that the% m / m concentration of that solution is 6.66%.

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a 25 ml volume of a sodium hydroxide solution requires 19.6 mL of a 0.189 M HCl acid for neutralization. A 10 mL volume of phosp
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Answer: 0.172 M

Explanation:

a) To calculate theconcentration of base, 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 which is NaOH.

We are given:

n_1=1\\M_1=0.189M\\V_1=19.6mL\\n_2=1\\M_2=?\\V_2=25mL

Putting values in above equation, we get:

1\times 0.189\times 19.6=1\times M_2\times 25\\\\M_2=0.148

b) To calculate the concentration 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 H_3PO_4

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

We are given:

n_1=3\\M_1=?\\V_1=10mL\\n_2=1\\M_2=0.148\\V_2=34.9mL

Putting values in above equation, we get:

3\times M_1\times 10=1\times 0.148\times 34.9\\\\M_1=0.172M

The concentration of the phosphoric acid solution is 1.172 M

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3 years ago
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