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A sample of nickel was dissolved in nitric acid to produce a solution with a volume of 50.00 mL - HSC - SSCE Chemistry - Question 14 - 2021 - Paper 1
Question 14
A sample of nickel was dissolved in nitric acid to produce a solution with a volume of 50.00 mL. 10.00 mL of this solution was then diluted to 250.0 mL. This solutio... show full transcript
Worked Solution & Example Answer:A sample of nickel was dissolved in nitric acid to produce a solution with a volume of 50.00 mL - HSC - SSCE Chemistry - Question 14 - 2021 - Paper 1
Step 1
Step 2
Calculate the total amount of nickel in the diluted solution
Answer
Using the formula:
[ C = \frac{n}{V} ]
Where:
- C = concentration in mol L⁻¹
- n = amount of substance in moles
- V = volume in liters
The volume of the diluted solution is 250.0 mL, which is 0.250 L. Therefore:
[ n = C \times V = 0.0015 , \text{mol L}^{-1} \times 0.250 , \text{L} = 0.000375 , \text{mol} ]
Step 3
Convert moles of nickel to mass
Answer
The molar mass of nickel (Ni) is approximately 58.69 g/mol. Thus:
[ m = n \times M = 0.000375 , \text{mol} \times 58.69 , \text{g/mol} \approx 0.0220 , ext{g} ]
Since 10.00 mL of the original 50.00 mL solution was taken for dilution, we need to scale this back:
[ m_{original} = 0.0220 , ext{g} \times \frac{50.00}{10.00} = 0.110 , ext{g} ]
However, with the choices, we see the closest available option is 0.15 g.