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Alborosie
3 years ago
11

At what temperature will 0.100 molal (M) NaCl(aq) boil? Kb= 0.51 C kg mol^-1

Chemistry
1 answer:
Talja [164]3 years ago
7 0
The increase of the boling point of a solution is a colligative property.

The formula for the increase of the normal boiling point of water is:

ΔTb = Kb * m

Where m is the molallity of the solution and Kb is the molal boiling constant in °C/mol.

ΔTb = 0.51 °C / m * 0.100 m = 0.051 °C.

So, the new boiling temperature is Tb = 100°C + 0.051°C = 100.051 °C.

Answer: 100.051 °C
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3) Calculate the percent by mass of 3.55 g NaCl dissolved in 88 g water.
kotykmax [81]

Answer:

The percent by mass of 3.55 g NaCl dissolved in 88 g water is 3.88%

Explanation:

When a solute dissolves in a solvent, the mass of the resulting solution is a sum of the mass of the solute and the solvent.

A percentage is a way of expressing a quantity as a fraction of 100. In this case, the percentage by mass of a solution is the number of grams of solute per 100 grams of solution and can be represented mathematically as:

Percent by mass=\frac{mass of solute}{mass of solution} *100

In this way it allows to precisely establish the concentration of solutions and express them in terms of percentages.

In this case:

  • mass of solute: 3.55 g
  • mass of solution: 3.55 g + 88 g= 91.55 g

Replacing:

Percentbymass=\frac{3.55}{91.55}*100

Percent by mass= 3.88%

<u><em>The percent by mass of 3.55 g NaCl dissolved in 88 g water is 3.88%</em></u>

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Answer:

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Explanation:

Accuracy is how close the value you gain from test or observation compared with the real value. In this test, all tests relatively close to 1kg and the average of the test is 1kg so this tool can be considered quite accurate.

The precision determines how close the value of multiple repetitions of the test. In this case, the value varies from 0.8kg to 1.2kg. The highest value is about 50% higher than the lowest value, so the range is pretty far and the test can be considered not precise.

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