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Tanya [424]
3 years ago
13

The accepted value for the percent by mass of water in a hydrate is 36.0%. In a laboratory activity, a student determined the pe

rcent by mass of water in the hydrate to be 37.8%. What is the percent error for the student's measured value?
(1) 5.0% (3) 1.8%

(2) 4.8% (4) 0.05%
Chemistry
1 answer:
valentina_108 [34]3 years ago
4 0
The original or accepted value for the percent by mass of water in a hydrate = 36%
Percen by mass of water in the hydrate determined by the student
in the laboratory = 37.8%
So the difference between the actual and the percent by mass in water determined by student = (37.8 - 36.0)%
             = 1.8%
So the percentage of error made by the student = (1.8/36) * 100 percent
                                                                             = (18/360) * 100 percent
                                                                             = ( 1/20) * 100 percent
                                                                              = 5 percent
So the student makes an error of 5%. Option "1" is the correct option.
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A bottle of wine contains 12.8% ethanol by volume. The density of ethanol (CH3OH) is 0.789 g/cm. Calculate the concentratic etha
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Answer : The mass percent of ethanol is, 10.38 % and the molality of ethanol is, 2.52 mole/kg

Explanation :

In wine, the solute and solvent are ethanol and water respectively.

Given :

12.8 % ethanol by volume means 12.8 mL ethanol present in 100 mL solution.

Volume of ethanol = 12.8 mL

Volume of solution = 100 mL

Volume of water = 100 - 12.8 = 87.2 mL

Density of ethanol = 0.789g/cm^3=0.789g/mL

Density of water = 1 g/mL

Now we have to calculate the mass of ethanol and water.

\text{Mass of ethanol}=\text{Density of ethanol}\times \text{Volume of ethanol}=0.789g/mL\times 12.8mL=10.1g

and,

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/mL\times 87.2mL=87.2g

Now we have to calculate the total mass of 100 mL of wine.

Total mass of 100 mL of wine = 10.1 + 87.2 = 97.3 g

Now we have to calculate the mass percent of ethanol.

\text{Mass of percent of ethanol}=\frac{\text{Mass of ethanol}}{\text{Total mass of solution}}\times 100

\text{Mass of percent of ethanol}=\frac{10.1g}{97.3g}\times 100=10.38\%

The mass percent of ethanol is, 10.38 %

Now we have to calculate the molality.

Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}

Molar mass of ethanol = 46 g/mole

Molality=\frac{10.1g\times 1000}{46g/mole\times 87.2g}=2.52mole/kg

The molality of ethanol is, 2.52 mole/kg

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