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IRISSAK [1]
4 years ago
11

A bottle of wine contains 12.8% ethanol by volume. The density of ethanol (CH3OH) is 0.789 g/cm. Calculate the concentratic etha

nol in wine in terms of mass percent and molality. Mass percent Molality =
Chemistry
1 answer:
Scrat [10]4 years ago
6 0

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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lutik1710 [3]

The freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is  - 0.45°C

Using the equation,

ΔT_{f} = iK_{f}m

where:

ΔT_{f} = change in freezing point (unknown)

i = Van't Hoff factor

K_{f} = freezing point depression constant

m = molal concentration of the solution

Molality is expressed as the number of moles of the solute per kilogram of the solvent.

Molal concentration is as follows;

MM KCl = 74.55 g/mol

molal concentration = \frac{5.0g*\frac{1mol}{74.55g} }{550.0g*\frac{1kg}{1000g} }

molal concentration = 0.1219m

Now, putting in the values to the equtaion ΔT_{f} = iK_{f}m we get,

ΔT_{f} = 2 × 1.86 × 0.1219

ΔT_{f} = 0.4536°C

So, ΔT_{f} of solution is,

ΔT_{f_{solution} } = 0.00°C - 0.45°C

ΔT_{f_{solution} } =  - 0.45°C

Therefore,freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is  - 0.45°C

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7 0
2 years ago
How many molecules of Na are in 3.2 moles of Na?
pav-90 [236]

Answer: 1.9 x 10²⁴ molecules Na

Explanation: To solve for the molecules of Na, we will use the Avogadro's number.

3.2 moles Na x 6.022 x10²³ molecules Na/ 1 mole Nà

= 1.9 x 10²⁴ molecules Na

7 0
3 years ago
1.24 moles of magnesium arsenate are dissolved in 1.74 kg of solution. Calculate the molality of the solution.
aleksklad [387]

Answer:

Molality of the solution = 0.7294 M

Explanation:

Given:

Number of magnesium arsenate = 1.24 moles

Mass of solution = 1.74 kg

Find:

Molality of the solution

Computation:

Molality of the solution = Mole of solute / Mass of solution = 1.74 kg

Molality of the solution = 1.24 / 1.7

Molality of the solution = 0.7294 M

6 0
3 years ago
quizlet a basement is successfully sealed so that more radon may not enter the space, but 5.7×107 radon atoms are already trappe
12345 [234]

After 25 days, it remains radon 5.9x10^5 atoms.

Half-life is the time required for a quantity (in this example number of radioactive radon) to reduce to half its initial value.

N(Ra) = 5.7×10^7; initial number of radon atoms

t1/2(Ra) = 3.8 days; the half-life of the radon is 3.8 days

n = 25 days / 3.8 days

n = 6.58; number of half-lifes of radon

N1(Ra) = N(Ra) x (1/2)^n

N1(Ra) = 5.7×10^7 x (1/2)^6.58

N1(Ra) = 5.9x10^5; number of radon atoms after 25 days

The half-life is independent of initial concentration (size of the sample).

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Write and balance the half-reaction for the oxidation of white phosphorous P4 to the phosphate ion PO3^−4 in a basic solution.
aalyn [17]

Since the half-reaction is occurring in a basic solution, add 32OH− to each side of the equation to eliminate the H+ ions.

P₄ +16H₂O + 32OH⁻ ⟶ 4PO₃⁻⁴ + 32H⁺ +32OH⁻

Final reaction :

P₄ + 32OH⁻ ⟶ 4PO₃⁻⁴ + 16H₂O + 20e⁻

A half reaction is either the oxidation or reduction reaction component of a redox reaction. A half reaction is obtained by considering the change in oxidation states of individual substances involved in the redox reaction.

The concept of half-reactions is used to describe what occurs in an electrochemical cell, such as a Galvanic cell battery. Half-reactions can be written to describe both the metal undergoing oxidation (known as the anode) and the metal undergoing reduction (known as the cathode).

Half-reactions are often used as a method of balancing redox reactions. For oxidation-reduction reactions in acidic conditions, after balancing the atoms and oxidation numbers, one will need to add H+ ions to balance the hydrogen ions in the half reaction.

For oxidation-reduction reactions in basic conditions, after balancing the atoms and oxidation numbers, first treat it as an acidic solution and then add OH- ions to balance the H+ ions in the half reactions (which would give H2O).

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3 0
2 years ago
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