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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Answer : The volume of oxygen at STP is 112.0665 L
Solution : Given,
The number of moles of 
= 5 moles
At STP, the temperature is 273 K and pressure is 1 atm.
Using ideal gas law equation :
where,
P = pressure of gas
V = volume of gas
n = the number of moles
T = temperature of gas
R = gas constant = 0.0821 L atm/mole K (Given)
By rearranging the above ideal gas law equation, we get
Now put all the given values in this expression, we get the value of volume.
Therefore, the volume of oxygen at STP is 112.0665 L
The answer is “Only some of the molecules of a weak base dissociate to produce hydroxide ions when mixed with water, but all of the molecules of a strong base dissociate to produce hydroxide ions”
