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3 years ago

What volume of a 0.33-M C12H22O11 solution can be diluted to prepare 25 mL of a solution with a concentration of 0.025 M?

Chemistry

1 answer:

Vsevolod [243]3 years ago

7 0

The dilution formula can be used to find the volume needed

c1v1 = c2v2

Where c1 is concentration and v1 is volume of the concentrated solution

And c2 is concentration and v2 is volume of the diluted solution to be prepared

c1 - 0.33 M

c2 - 0.025 M

v2 - 25 mL

Substituting these values in the equation

0.33 M x v1 = 0.025 M x 25 mL

v1 = 1.89 mL

Therefore 1.89 mL of the 0.33 M solution needs to be diluted up to 25 mL to make a 0.025 M solution

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2 years ago

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3 years ago

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3 years ago

It was calculated that 4.3mL of 0.417 M HCl is required to titrate 11.9 mL of 0.151 M Mg(OH)2. Show evidence 2 HCl(aq) + Mg(OH)2

Lapatulllka [165]

Answer:

See explanation.

Explanation:

Hello,

In this case, for the described chemical reaction:

2 HCl(aq) + Mg(OH)2(aq) → MgCl2(aq) + 2 H2O(l)

We can notice there is a 2:1 molar ratio between the moles of hydrochloric acid and magnesium hydroxide, therefore, at the equivalence point:

$n_{HCl}=2*n_{Mg(OH)_2}$

And in terms of volumes and concentrations we verify:

$V_{HCl}M_{HCl}=2*V_{Mg(OH)_2}M_{Mg(OH)_2}$

So we use the given data to proof it:

$4.3mL*0.417M=2*11.9mL*0.151M\\1.793=3.594$

Therefore, we can conclude the data is wrong by means of the 2:1 mole ratio that for sure was not taken into account. This is also supported by the fact that normalities are actually the same, but the nomality of magnesium hydroxide is the half of the hydrochloric acid normality since the acid is monoprotic and the base has two hydroxyl ions.

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