The effective molar mass of ashes has units of grams of ashes per mole of base provided. The mass of the ashes in the experiment
1 answer:
Answer:
molar mass of unknown monoprotic acid = 114.1 g / mole.
Explanation:
Note: This question is incomplete and lacks necessary data to solve for the required calculation. However, I have similar question on the internet and seen the question completely and will be using that data to solve for this question in order to solve required calculation. Besides that, complete question asks us to solve for molar mass of the acid used.
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Calculation:
Volume = 34.81mL
34.81 mL of 0.4346 M potassium hydroxide
= 0.03481 L * 0.4346 mole / L
= 0.01513 mole.
Balance equation:
HA + KOH ----> KA + H2O
As we know from the question,
one mole HA neutralize with 1 mole KOH.
mole of unknown monoprotic acid = 0.01513 mole.
And we know that the formula for mole is:
mole = mass / molar mass
Making molar mass as the subject:
Molar mass = mass / mole
Molar mass = 1.726 g / 0.01513 mole
Molar mass = 114.1 g / mole.
So,
Molar mass of unknown monoprotic acid = 114.1 g / mole.
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<u>Answer:</u> The concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.
<u>Explanation:</u>
To calculate the number of moles for given molarity, we use the equation:
.....(1)
<u>For NaOH:</u>
Initial molarity of NaOH solution = 0.195 M
Volume of solution = 15.0 mL = 0.015 L (Conversion factor: 1 L = 1000 mL)
Putting values in equation 1, we get:
<u>For sulfuric acid:</u>
Initial molarity of sulfuric acid solution = 0.248 M
Volume of solution = 20.0 mL = 0.020 L
Putting values in equation 1, we get:
The chemical equation for the reaction of NaOH and sulfuric acid follows:
By Stoichiometry of the reaction:
2 moles of NaOH reacts with 1 mole of sulfuric acid
So, moles of KOH will react with =
of sulfuric acid
As, given amount of sulfuric acid is more than the required amount. So, it is considered as an excess reagent.
Thus, NaOH is considered as a limiting reagent because it limits the formation of product.
Excess moles of sulfuric acid =
By Stoichiometry of the reaction:
2 moles of KOH produces 1 mole of sodium sulfate
So, moles of KOH will produce =
of sodium sulfate
- <u>For sodium sulfate:</u>
Moles of sodium sulfate =
Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L
Putting values in equation 1, we get:
- <u>For sulfuric acid:</u>
Moles of excess sulfuric acid =
Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L
Putting values in equation 1, we get:
- <u>For NaOH:</u>
Moles of NaOH remained = 0 moles
Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L
Putting values in equation 1, we get:
Hence, the concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.