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IRISSAK [1]
2 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]2 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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OLga [1]
Let's assume that the gas has ideal gas behavior.

Then we can use ideal gas equation,
PV = nRT

Where, <span>
P = Pressure of the gas (Pa)
V = volume of the gas (m³)
n = number of moles (mol)
R = Universal gas constant (8.314 J mol</span>⁻¹ K⁻¹)<span>
T = temperature in Kelvin (K)
<span>
The given data for the </span></span>gas is,<span>
P = 2.8 atm = 283710 Pa
V = 98 L = 98 x 10</span>⁻³ m³<span>
T = 292 K
R = 8.314 J mol</span>⁻¹ K⁻¹<span>
n = ?

By applying the formula,
283710 Pa x </span>98 x 10⁻³ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 292 K
<span>                                       n = 11.45 mol

Hence, moles of gas is </span>11.45 mol.
4 0
2 years ago
Use the information in the square to answer the questions about copper. A purple box has C u at the center and 29 above. Below i
scoray [572]

Answer:

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6 0
2 years ago
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OverLord2011 [107]

Answer : The number of moles of oxygen present in a sample are 11.3 moles.

Explanation :

The given compound is, Ni(CO)_4

By the stoichiometry we can say that, 1 mole of of Ni(CO)_4 has 4 moles of CO.

Or we can say that, 1 mole of of Ni(CO)_4 has 1 mole of nickel (Ni), 4 moles of carbon (C) and 4 moles of oxygen.

That means,

Number of moles of carbon = Number of moles of oxygen

As we are given that:

Number of moles of carbon = 11.3 moles

So, number of moles of oxygen = number of moles of carbon = 11.3 moles

Therefore, the number of moles of oxygen present in a sample are 11.3 moles.

7 0
2 years ago
The decomposition of nitramide, O 2 NNH 2 , O2NNH2, in water has the chemical equation and rate law O 2 NNH 2 ( aq ) ⟶ N 2 O ( g
valkas [14]

Answer:

Explanation:

The given overall reaction is as follows:

O 2 N N H ₂( a q ) k → N ₂O ( g ) + H ₂ O( l )

The reaction mechanism for this reaction is as follows:

O ₂ N N H ₂ ⇌ k 1 k − 1  O ₂N N H ⁻ + H ⁺ ( f a s t  e q u i l i b r i u m )

O ₂ N N H − k ₂→ N ₂ O + O H ⁻ ( s l ow )

H ⁺ + O H − k ₃→ H ₂ O ( f a s t )

The rate law of the reaction is given as follows:

k = [ O ₂ N N H ₂ ]  / [ H ⁺ ]

The rate law can be determined by the slow step of the mechanism.

r a t e = k ₂ [ O ₂ N N H ⁻ ] . . . ( 1 )

Since, from the equilibrium reaction

k e q = [ O ₂ N N H ⁻ ] [ H ⁺ ] /[ O ₂ N N H ₂ ] = k ₁ /k − 1

[ O ₂ N N H ⁻] = k ₁ /k − 1  × [ O ₂ N N H ₂ ] /[ H ⁺ ]. . . . ( 2 )

Substitituting the value of equation (2) in equation (1) we get.

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Therefore, the overall rate constant is

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5 0
1 year ago
A metal oxide with the formula mo contains 15.44% oxygen. in the box below, type the symbol for the element represented by m.
lana66690 [7]

<u>Answer:</u> The element represented by M is Strontium.

<u>Explanation:</u>

Let us consider the molar mass of metal be 'x'.

The molar mass of MO will be = Molar mass of oxygen + Molar mass of metal = (16 + x)g/mol

It is given in the question that 15.44% of oxygen is present in metal oxide. So, the equation becomes:

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The metal atom having molar mass as 87.62/mol is Strontium.

Hence, the element represented by M is Strontium.

8 0
1 year ago
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