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mezya [45]
3 years ago
8

Determine molar concentration of the equivalent of acetic acid in 6% solution (d=1g/ml)

Chemistry
1 answer:
Aleksandr-060686 [28]3 years ago
8 0

Molar concentration of acetic acid = 1 M

<h3>Further explanation</h3>

Given

acetic acid in 6% solution (d=1 g/ml)

Required

molar concentration

Solution

mass of acetic acid in solution

mass = density x %mass

mass = 1 g/ml x 6%

mass = 0.06 g/ml ⇒ 60 g/L

MW acetic acid = 60 g/mol

molarity acetic acid = mass : MW

molarity = 60 g/L : 60 g/mol

molarity = 1 mol / L = 1 M

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Calculate the pH of a buffer solution made by adding 15.0 g anhydrous sodium acetate (NaC2H3O2) to 100.0 mL of 0.200 M acetic ac
DiKsa [7]

Answer:

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Explanation:

Mole of anhydrous sodium acetate = \frac{Given mass}{Molecular mass}

                                                           = \frac{15}{82}

                                                           = 0.18 mole

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= 20 mmol

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So pH of the Buffer solution = 5.69

8 0
3 years ago
What is the molarity of a solution made by dissolving 18.9g of ammonium nitrate in enough water to make 855 ml of solution?
Leya [2.2K]
Answer is: the molarity of a solution is 0.276 M.<span>

V(solution) = 855 mL </span>÷ 1000 mL/L.
V(solution) = 0.855 L.
m(NH₄NO₃) = 18.9 g; mass of ammonium nitrate.
M(NH₄NO₃) = 80.04 g/mol; molar mass of ammonium nitrate.
n(NH₄NO₃) = m(NH₄NO₃) ÷ M(NH₄NO₃).
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7 0
3 years ago
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Answer:

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3 0
3 years ago
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3 years ago
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