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Elden [556K]
3 years ago
12

Calculate the concentration of A bottle of wine contains 12.9% ethanol by volume. The density of ethanol (CH3OH) is 0.789 g/cm e

thanol in wine in terms of mass percent and molality Mass percent Molality =
Chemistry
1 answer:
Delicious77 [7]3 years ago
4 0

Answer:

The mass percentage of the solution is 10.46%.

The molality of the solution is 2.5403 mol/kg.

Explanation:

A bottle of wine contains 12.9% ethanol by volume.

This means that in 100 mL of solution 12.9  L of alcohol is present.

Volume of alcohol = v = 12.9 L

Mass of the ethanol = m

Density of the ethanol ,d= 0.789 g/cm^3=0.789 g/mL

1 cm^3=1 mL

m=d\times v=0.798 g/ml\times 12.9 mL = 10.1781 g

Mass of water = M

Volume of water ,V= 100 mL - 12.9 mL = 87.1 mL

Density of water = D=1.00 g/mL

M=D\times V=1.00 g/ml\times 87.1 mL =87.1 g

Mass percent

(w/w)\%=\frac{m}{m+M}\times 100

\frac{10.1781 g}{10.1781 g+87.1 g}\times 100=10.46\%

Molality :

m=\frac{m}{\text{molar mass of ethanol}\times M(kg)}

M = 87.1 g = 0.0871 kg (1 kg =1000 g)

=\frac{10.1781 g}{46 g/mol\times 0.0871 kg}

m=2.5403 mol/kg

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6 0
2 years ago
Calculate the number of atoms in 2.0g of hydrogen atoms?
Jobisdone [24]
First you calculate how many moles there are in 2.0 grams of hydrogen (H2) atoms.
Hydrogen has a relative atomic mass (RAM) of 1 g/mol, but there are 2 hydrogen atoms: 1 x 2 = 2 g/mol

To work out how many moles there are,
use the formula: n(moles) = mass ÷ molar mass
n(moles) = 2 grams ÷ 2 g/mol = 1 mol

Then use Avogadro's Constant : 6.023 x 10^23
= 1 x 6.023 x 10^23
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Final step is to multiply it by the number of atoms, in this case there are 2.
= 6.023 x 10^23 x 2
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4 0
3 years ago
A balloon containing helium gas expands from 230
Anit [1.1K]

The answer for the following problem is mentioned below.

  • <u><em>Therefore the final  moles of the gas is 14.2 × </em></u>10^{-4}<u><em> moles.</em></u>

Explanation:

Given:

Initial volume (V_{1}) = 230 ml

Final volume (V_{2}) = 860 ml

Initial moles (n_{1}) = 3.8 ×10^{-4} moles

To find:

Final moles (n_{2})

We know;

According to the ideal gas equation;

    P × V = n × R × T

where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of the moles of the gas

R represents the universal gas constant

T represents the temperature of the gas

So;

    V ∝ n

\frac{V_{1} }{V_{2} } = \frac{n_{1} }{n_{2} }

where,

(V_{1}) represents the initial volume of the gas

(V_{2}) represents the final volume of the gas

(n_{1}) represents the initial  moles of the gas

(n_{2}) represents the final moles of the gas

Substituting the above values;

   \frac{230}{860} = \frac{3.8 * 10^-4}{n_{2} }

  n_{2} = 14.2 × 10^{-4} moles

<u><em>Therefore the final  moles of the gas is 14.2 × </em></u>10^{-4}<u><em> moles.</em></u>

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Answer:

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