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belka [17]
3 years ago
13

An alloy of bronze is manufactured by melting 51.2 g of copper with 6.84 g of tin. What is the percent copper in the bronze?

Chemistry
1 answer:
Harlamova29_29 [7]3 years ago
5 0

Answer:

Percentage of copper = 88%

Explanation:

Given data:

Mass of copper = 51.2 g

Mass of tin = 6.84 g

Percentage of copper = ?

Solution:

Formula:

Percentage of copper = mass of copper / total mass × 100

Now we will determine the total mass:

Total mass = mass of copper + mass of  tin

Total mass = 51.2 g + 6.84 g

Total mass = 58.04 g

Now we will calculate the percentage of copper.

Percentage of copper = 51.2 g / 58.04 g × 100

Percentage of copper = 0.88 × 100

Percentage of copper = 88%

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Norfenefrine (C₈H₁₁NO₂).

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We will solve a case related to one of the colligative properties, namely freezing point depression.

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\boxed{ \ \Delta T_f = T_f(solvent) - T_f(solution) \ } \rightarrow \boxed{ \ \Delta T_f = K_f \times molality \ }

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A mysterious white powder could be,

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When 82 mg of the powder is dissolved in 1.50 mL of ethanol (density = 0.789 g/cm³, normal freezing point −114.6°C, Kf = 1.99°C/m), the freezing point is lowered to −115.5°C.

<u>Question: </u>What is the identity of the white powder?

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Let us identify the solute, the solvent, initial, and final temperatures.

  • The solute = the powder
  • The solvent = ethanol
  • The freezing point of the solvent = −114.6°C
  • The freezing point of the solution = −115.5°C

Prepare masses of solutes and solvents.

  • Mass of solute = 82 mg = 0.082 g
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We must prepare the solvent mass unit in kg because the unit of molality is the mole of the solute divided by the mass of the solvent in kg.

The molality formula is as follows:

\boxed{ \ m = \frac{moles \ of \ solute}{kg \ of \ solvent} \ } \rightarrow \boxed{ \ m = \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }

Now we combine it with the formula of freezing point depression.

\boxed{ \ \Delta T_f =  K_f \times \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }

It is clear that we will determine the molar mass of the solute (denoted by Mr).

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\boxed{ \ -114.6^0C - (-115.5^0C) = 1.99 \frac{^0C}{m} \times \frac{0.082 \ g}{Mr \times 0.00118 \ kg} \ }

\boxed{ \ 0.9 = \frac{1.99 \times 0.082}{Mr \times 0.00118} \ }

\boxed{ \ Mr = \frac{0.16318}{0.9 \times 0.00118} \ }

We get \boxed{ \ Mr = 153.65 \ }

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