A food scientist also tested some ethanoic acid solution. Ethanoic acid is the acid which is found in vinegar.His results are shown below.
vinegar ethanoic
acid
Volume used/Cm3 50 10
volume of NaOH needed to neutralise/cm 30 10
The ethanoic acid solution contained 10 g of acid in every 100 cm3 of solution.
(b) (i) How many grams of ethanoic acid are there in 10 cm3 of solution ?
(ii) Work out how many grams of ethanoic acid are in 50 cm3 of the vinegar.
To answer these questions, we need to use the given information about the volume of ethanoic acid solution and the concentration of ethanoic acid in the solution.
(i) How many grams of ethanoic acid are there in 10 cm3 of solution?
First, we know that the ethanoic acid solution contains 10 g of acid in every 100 cm3 of solution. So, to find the grams of ethanoic acid in 10 cm3, we can use the proportion:
10 g / 100 cm3 = x g / 10 cm3
Cross-multiplying, we can find the value of x:
10 cm3 * 10 g = 100 cm3 * x g
100 g = 100 cm3 * x g
Dividing both sides by 100 cm3, we get:
x g = 100 g / 100
x g = 1 g
Therefore, there are 1 gram of ethanoic acid in 10 cm3 of solution.
(ii) Work out how many grams of ethanoic acid are in 50 cm3 of the vinegar.
For this question, we are given the volume of vinegar used (50 cm3) and the volume of NaOH needed to neutralize it (30 cm3). From this information, we can determine the volume of ethanoic acid.
Since 10 cm3 of ethanoic acid solution required 10 cm3 of NaOH to neutralize, we can set up a proportion:
10 cm3 / 30 cm3 = 50 cm3 / x cm3
Cross-multiplying, we get:
10 cm3 * x cm3 = 30 cm3 * 50 cm3
x cm3 = (30 cm3 * 50 cm3) / 10 cm3
Simplifying the equation, we find:
x cm3 = 150 cm3
Therefore, the volume of ethanoic acid in 50 cm3 of vinegar is 150 cm3.
Lastly, to determine the grams of ethanoic acid in 50 cm3 of vinegar, we can again use the given concentration of the acid:
10 g / 100 cm3 = x g / 150 cm3
Cross-multiplying, we find:
150 cm3 * 10 g = 100 cm3 * x g
1500 g = 100 cm3 * x g
Dividing both sides by 100 cm3, we get:
x g = 1500 g / 100
x g = 15 g
Therefore, there are 15 grams of ethanoic acid in 50 cm3 of vinegar.