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
molar mass of water = 18 g/mol
moles of vinegar = 5/62 = 0.08
moles of water = 95/18 = 5.28
total moles = 5.36
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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The term that refers to compounds that can form hydrates but do not contain water molecules is anhydrous.
Gravity is the force that draws water back to the earth in the forms of rain, sleet and, snow.
Answer:
2.64%
Explanation:
mass percent = (grams of solute / grams of solution) x 100
mass percent = (2.4 / 91) × 100
mass percent = 2.64% to 3sf
Answer:
Molarity = 2.3 M
Explanation:
Molarity can be calculated using the following rule:
Molarity = number of moles of solute / volume of solution
1- getting the number of moles:
We are given that:
mass of solute = 105.96 grams
From the periodic table:
atomic mass of carbon = 12 grams
atomic mass of hydrogen = 1 gram
atomic mass of oxygen = 16 grams
Therefore:
molar mass of C2H6O = 2(12) + 6(1) + 16 = 46 grams
Now, we can get the number of moles as follows:
number of moles = mass / molar mass = 105.96 / 46 = 2.3 moles
2- The volume of solution is given = 1 liter
3- getting the molarity:
molarity = number of moles of solute / volume of solution
molarity = 2.3 / 1
molarity = 2.3 M
Hope this helps :)
