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serg [7]
3 years ago
7

Vinegar is a solution of acetic acid (the solute) in water (the solvent) with a solution density of 1010 g/L. If vinegar is 0.80

M acetic acid, what is the % by mass concentration of acetic acid in vinegar?
Chemistry
1 answer:
Damm [24]3 years ago
6 0

Answer:

4.8 %

Explanation:

We are asked the concentration in % by mass, given the molarity of the solution and its density.

0.8 molar solution means that we have 0.80 moles of acetic acid in 1 liter of solution. If we convert the moles of acetic acid to grams, and the 1 liter solution to grams, since we are given the density of solution, we will have the values necessary to calculate the % by mass:

MW acetic acid = 60.0 g/mol

mass acetic acid (the solute) = 0.80 mol x 60 g / mol = 48.00 g

mass of solution = 1000 cm³ x 1.010 g/ cm³      (1l= 1000 cm³)

                            = 1010 g

% (by mass) = 48.00 g/ 1010 g  x 100 = 4.8 %

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The problem can solved using the heat equation which is expressed as:

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A perfectly spherical balloon occupied 35°C of neon gas under a pressure of 2 atm.
lutik1710 [3]

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Explanation:

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Answer:

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Hello,

In this case, the molarity is defined as the ratio of the moles of the solute to the volume of the solution in liters:

M=\frac{n_{solute}}{V_{solution}}

In this case, the solute is the KCl (potassium chloride) and the solution is made up of both water and KCl. Moreover, since during this type of dissolution processes, the volume of the solution is not significantly affected by the addition of the solute, the resulting molarity is:

M=\frac{0.88mol}{2.0L}\\ \\M=0.44M

Best regards.

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