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Explanation:
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<u>Answer:</u> The molality of potassium hydroxide solution is 0.608 m
<u>Explanation:</u>
We are given:
3.301 mass % of potassium hydroxide solution.
This means that 3.301 grams of potassium hydroxide is present in 100 grams of solution
Mass of solvent = Mass of solution - Mass of solute (KOH)
Mass of solvent = (100 - 3.301) g = 96.699 g
To calculate the molality of solution, we use the equation:
Where,
= Given mass of solute (KOH) = 3.301 g
= Molar mass of solute (KOH) = 56.1 g/mol
= Mass of solvent = 96.699 g
Putting values in above equation, we get:
Hence, the molality of potassium hydroxide solution is 0.608 m
Answer: A) More base is likely required to reach the endpoint for the diprotic acid than for the monoprotic acid under these conditions
Explanation:
The monoprotic acid (HA) has a valency of 1 and diprotic acid has a valency of 2.
As the concentration and volume of the diprotic acid and the monoprotic acids are equal.
The neutralization reaction for monoprotic acid is:
The neutralization reaction for diprotic acid is:
Thus more number of moles of base are required for neutralization of diprotic acid and thus the volume required will be more as concentration and volume of the diprotic acid and the monoprotic acids are equal.
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Test tube of ammonium chloride (NH4Cl) being heated over a bunsen burner flame. Ammonium chloride decomposes readily when heated, but condenses in the cooler area at the top of the test tube. This is a reversible reaction, where the ammonium chloride decomposes into the gases ammonia (NH3) and hydrogen chloride (HCl).