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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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Answer:

The correct answer is Pu, 234.

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In the given case, let us consider the reactant as X. Now the mass number (balanced) on both the sides will be,

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M = 1 * 103 + 1 * 131 + 2 * 0

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Now the atomic number (balanced) on both the sides,

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A = 1*42 + 1*50 + 2*1

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Substituting the values,

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