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Question 6 (6 marks) Brass is an alloy of copper and zinc: To determine the percentage of copper in a particular sample of brass, an analyst prepared a number of standard solutions of copper(II) ions and measured their absorbance using an atomic absorption spectrometer (AAS) - VCE - SSCE Chemistry - Question 6 - 2016 - Paper 1

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Question 6

Question-6-(6-marks)--Brass-is-an-alloy-of-copper-and-zinc:-To-determine-the-percentage-of-copper-in-a-particular-sample-of-brass,-an-analyst-prepared-a-number-of-standard-solutions-of-copper(II)-ions-and-measured-their-absorbance-using-an-atomic-absorption-spectrometer-(AAS)-VCE-SSCE Chemistry-Question 6-2016-Paper 1.png

Question 6 (6 marks) Brass is an alloy of copper and zinc: To determine the percentage of copper in a particular sample of brass, an analyst prepared a number of st... show full transcript

Worked Solution & Example Answer:Question 6 (6 marks) Brass is an alloy of copper and zinc: To determine the percentage of copper in a particular sample of brass, an analyst prepared a number of standard solutions of copper(II) ions and measured their absorbance using an atomic absorption spectrometer (AAS) - VCE - SSCE Chemistry - Question 6 - 2016 - Paper 1

Step 1

A. Calculate c(Cu²⁺) from absorbance

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Answer

Using the calibration curve, the absorbance of 0.13 corresponds to a concentration of copper(II) ions (c(Cu²⁺)).

From the marking scheme, the linear relationship gives c(Cu2+)=1.1gL1c(Cu^{2+}) = 1.1 \, g \, L^{-1}.

Step 2

B. Calculate m(Cu) in 100 mL solution

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Answer

To find the mass of copper in the 100 mL solution, we use the formula:

m(Cu)=c(Cu2+)×Vm(Cu) = c(Cu^{2+}) \times V

Where:

  • c(Cu2+)=0.11gL1c(Cu^{2+}) = 0.11 g \, L^{-1}
  • V=0.100LV = 0.100 \, L (100 mL)

So,

m(Cu)=0.11gL1×0.100L=0.011gm(Cu) = 0.11 \, g \, L^{-1} \times 0.100 \, L = 0.011 \, g

Step 3

C. Calculate percentage of copper in the brass sample

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Answer

The percentage by mass of copper in the brass sample can be calculated as follows:

%Cu=(m(Cu)msample)×100\% \text{Cu} = \left( \frac{m(Cu)}{m_{sample}} \right) \times 100

Here, msample=0.198gm_{sample} = 0.198 \, g.

Substituting values: %Cu=(0.0110.198)×1005.56%\% \text{Cu} = \left( \frac{0.011}{0.198} \right) \times 100 \approx 5.56\%

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