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yanalaym [24]
3 years ago
11

A 20.0 mL sample of a NaCl solution has a mass of 21.040g. After the solution is evaporated to dryness, the dry salt residue has

a mass of 4.348g. Calculate the following concentrations for the NaCl solution. (Molarity)
Chemistry
1 answer:
Vilka [71]3 years ago
7 0
Mass NaCl = 4.348
Mass solution = 21.040 - 4.348 = 16.692
mass/mass % =
4.348/16.692
= 26%
mass/vol % =
21.7%
Molecular mass of NaCl = 58.5
Moles of NaCl = 0.074
Volume of solution = 0.02 L
Molarity = 3.7 mol/Liter
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The mole ratio of acetic acid to water in 100 g of vinegar is 0.015 : 0.985.

<h3>What is the mole ratio of acetic acid to water in 100 g of vinegar?</h3>

The mole ratio of acetic acid to water in 100 g of vinegar is determined from their percentage composition.

The percentage composition of acetic acid and water in vinegar is 5% acetic acid and 95% water.

In 100 g of vinegar, there are 5 g of acetic acid and 5 g of water.

Moles = mass/molar mass

molar mass of acetic acid = 62 g/mol

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Mole ratio of vinegar to water = 0.08/5.36 : 5.28/5.36

Mole ratio of vinegar to water = 0.015 : 0.985

In conclusion, the mole ratio is determined from the percentage composition of acetic acid and water in vinegar.

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