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Semenov [28]
3 years ago
8

Brass is a substitutional alloy consisting of a solution of copper and zinc. A particular sample of red brass consisting of 79.0

% Cu and 21.0% Zn by mass has a density of 8740 kg/m3.What is the molality and molarity of Zn in the solid solution?
Chemistry
1 answer:
Bas_tet [7]3 years ago
4 0

<u>Answer:</u> The molality and molarity of zinc in the solution is 4.07 m and 28.11 M respectively.

<u>Explanation:</u>

Solute is the substance which is present in smaller proportion in a mixture and solvent is the substance which is present in larger proportion in a mixture.

We are given:

(m/m) % of Cu = 79 %

This means that 79 g of copper is present in 100 grams of alloy.

(m/m) % of Zn = 21 %

This means that 21 g of zinc is present in 100 grams of alloy.

As, zinc is present in smaller proportion. So, it is solute and copper is the solvent.

  • <u>Calculating molality of zinc:</u>

To calculate molality of the zinc, we use the equation:

Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

where,

m_{solute} = Given mass of solute (Zinc ) = 21 g

M_{solute} = Molar mass of solute (Zinc) = 65.3 g/mol

W_{solvent} = Mass of solvent (copper) = 79 g

Putting values in above equation, we get:

\text{Molality of Zinc}=\frac{21\times 1000}{65.3\times 79}\\\\\text{Molality of zinc}=4.07m

  • <u>Calculating molarity of zinc:</u>

To calculate volume of a substance, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of solution = 8740kg/m^3=\frac{8740kg\frac{1000g}{1kg}}{1m^3\times \frac{10^6cm^3}{1m^3}}=\frac{8740g}{1000cm^3}=8.740g/cm^3

(Conversion factors used are:  1 kg = 1000 g &  1m^3=10^6cm^3  )

Mass of solution = 100 g

Putting values in above equation, we get:

8.740g/cm^3=\frac{100.0g}{\text{Volume of zinc}}\\\\\text{Volume of zinc}=11.44cm^3

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Molar mass of zinc = 65.3 g/mol

Volume of solution = 11.44cm^3=11.44mL     (Conversion factor:  1cm^3=1mL  )

Mass of zinc = 21.0 g

Putting values in above equation, we get:

\text{Molarity of zinc}=\frac{21\times 1000}{65.3\times 11.44}=28.11M

Hence, the molality and molarity of zinc in the solution is 4.07 m and 28.11 M respectively.

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