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Dmitrij [34]
3 years ago
9

A student used water as the solvent and encountered some problems. Comment on the effect, if any, each of the following situatio

ns could have had on the experimental results...a)The unknown, a white powder, failed to dissolve.b)The student returned to the laboratory instructor for a different solid unknown. This unknown dissolved, but bubbles were seen escaping from the solution almost immediately after the addition of the solid.c)As the student was setting up the apparatus to measure the freezing point of the unknown solution, the thermometer assembly rolled off the lab bench, and the thermometer broke. The student observed a new thermometer and performed the experiment as instructed.
Chemistry
1 answer:
Troyanec [42]3 years ago
8 0

Answer:

a) If the white powder didn't dissolve completely, that could have a great effect on the experiment. The concentration of the solution cant get to the point where it needs to get for the rest of the experiment which can skew results.

b) If a solution forms bubbles immediately after an addition of a solid, that could simply mean that gas is formed in that reaction. That has no negative effect on an experiment since that's what is supposed to happen during the reaction.

c) Using a different measuring device can really effect a students calculations later on during the experiment. Different devices are calibrated differently, which can skew the results or calculations later on. Its best to stay consistent with what you are measuring with during an experiment.

Explanation:

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Answer : The volume for 6.0m HCl solution required = 62.5 ml

Solution : Given,

Initial concentration of HCl solution = 6.0m

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Initial volume of HCl solution = ?

Formula used for dilution is,

M_1V_1=M_2V_2

where,

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Now put all the given values in above formula, we get the initial volume of HCl solution.

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Therefore, the volume for 6.0m HCl solution required = 62.5 ml


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