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Nezavi [6.7K]
3 years ago
11

A 12.0% sucrose solution by mass has a density of 1.05 gem, what mass of sucrose is present in a 32.0-mL sample of this solution

?
A) 0.126g
B) 3.66g
C) 4.03g
D) 3.84g
E) 280 g
Chemistry
1 answer:
natulia [17]3 years ago
7 0

Answer:

Option C. 4.03 g

Explanation:

Firstly we analyse data.

12 % by mass, is a sort of concentration. It indicates that in 100 g of SOLUTION, we have 12 g of SOLUTE.

Density is the data that indicates grams of solution in volume of solution.

We need to determine, the volume of solution for the concentration

Density = mass / volume

1.05 g/mL = 100 g / volume

Volume =  100 g / 1.05 g/mL → 95.24 mL

Therefore our 12 g of solute are contained in 95.24 mL

Let's finish this by a rule of three.

95.24 mL contain 12 g of sucrose

Our sample of 32 mL may contain ( 32 . 12) / 95.24 = 4.03 g

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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

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\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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