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hammer [34]
3 years ago
6

What would be the effect of each of the following sources of error on the molarity of H2SO4 determined in this experiment? State

whether the concentration of H2SO4 obtained would be too high or too low. Explain your answer briefly.
a. The burette is contaminated with an acid solution.
[1 mark]

•

b. The burette contains a large air bubble in the tip, which disappears during the titration [1mark]

c. A small volume of the acid is spilled when you transfer it into the Erlenmeyer flask
[1 mark]
Chemistry
1 answer:
Dafna11 [192]3 years ago
7 0

Answer:

The burette is contaminated with an acid solution- The concentration is too high

The burette contains a large air bubble in the tip, which disappears during the titration- The concentration of the acid is too low.

A small volume of the acid is spilled when you transfer it into the Erlenmeyer flask- The concentration of the acid is too low.

Explanation:

The molality of a solution is the number of moles of solute per liter of solution. The concentration of an unknown solution is obtained by titration against a standard solution of acid or base whose concentration  is known. It is a volumetric method of analysis.

If the burette is contaminated with the acid, then the concentration of the base that reacts with the acid is less than the stated amount and consequently the concentration of the acid calculated is higher than it should be.

If the tip of the burette contains air bubbles, the volume of the acid reported is less than the volume of acid actually delivered during the titration hence the calculated concentration of the acid is too low.

If some volume of acid is spilled when you transfer the acid into the Erlenmeyer flask, the volume of the acid reacted is decreased and consequently the calculated concentration of the acid is too low.

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a sample of the hydrate of sodium carbonate has a mass of 8.85 g. it loses 1.28 g when heated. find the molar ratio of this comp
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Answer:

Molar ratio of the compound is 1:1 and the type of hydrate is Mono hydrate.

Explanation:

From the given,

Mass of sodium carbonate Na_{2}CO_{3}.XH_{2}O = 8.85 g

Loss mass H_{2}O = 1.28 g

Actual weight of sodium carbonate = 8.85 g - 1.28 g = 7.57 g

7.57 g Na_{2}CO_{3} \times \frac{1mol}{106 g} =\frac{0.0714}{0.0714}=1

1.28g H_{2}O \times \frac{1mol}{18 g} =\frac{0.0711}{0.0714}=1

Therefore, the compound has only one water molecule.

Molecular formula of the compound is Na_{2}CO_{3}.H_{2}O an name of the compound is <u>sodium carbonate mono hydrate.</u>

Hence, the type of the compound is Mono hydrate.

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