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mel-nik [20]
3 years ago
15

A piece of gold jewelry weighs 11.54 g and has a volume of 0.675 cm3. The jewelry contains only gold (density = 19.3 g/cm3) and

silver (density = 10.5 g/cm3). Assuming that the total volume of the jewelry is the sum of the volumes of the gold and silver that it contains, calculate the percentage of gold in the jewelry.
Chemistry
1 answer:
lbvjy [14]3 years ago
4 0

Answer:

Percentage of gold = 84.6%

Explanation:

Suppose the jewelry has x g of gold and y g of silver

The mass of both is the total mass of the jewelry then:

x + y = 11.54

And as the problem says the total volume is the sum of both volumes.

volume = mass / density

\frac{x}{19.3} + \frac{y}{10.5} = 0.675

Solving the system you get:

x = 9.764 g

y = 1.776 g

The percentage of gold in the jewelry is:

X_{gold} = \frac{9.764}{11.54} * 100 = 84.6\%

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V = Volume of the gas = 2.19 L

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Putting values in above equation, we get:

5.57bar\times 2.19L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{5.57\times 2.19}{0.0831\times 298}=0.493mol

To calculate the mole fraction of carbon dioxide, we use the equation given by Raoult's law, which is:

p_{A}=p_T\times \chi_{A}         ........(1)

where,

p_A = partial pressure of carbon dioxide = 0.318 bar

p_T = total pressure = 5.57 bar

\chi_A = mole fraction of carbon dioxide = ?

Putting values in above equation, we get:

0.318bar=5.57bar\times \chi_{CO_2}\\\\\chi_{CO_2}=\frac{0.381}{5.57}=0.0571

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\chi_A=\frac{n_A}{n_A+n_B}

We are given:

Moles of nitrogen gas = 0.221 moles

Mole fraction of nitrogen gas, \chi_{N_2}=\frac{0.221}{0.493}=0.448

Calculating the partial pressure of oxygen gas by using equation 1, we get:

Mole fraction of oxygen gas = (1 - 0.0571 - 0.448) = 0.4949

Total pressure of the system = 5.57 bar

Putting values in equation 1, we get:

p_{O_2}=5.57bar\times 0.4949\\\\p_{O_2}=2.76bar

Hence, the partial pressure of oxygen gas is 2.76 bar

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hope this helps!
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