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

Using the formula:

$n = C \times V$

where:

• (C = 0.0025 , \text{mol L}^{-1})
• (V = 0.250 , \text{L})

we have:

$n = 0.0025 \, \text{mol L}^{-1} \times 0.250 \, \text{L} = 0.000625 \, \text{mol}$

Since this concentration corresponds to the diluted solution that was obtained from the initial 10.00 mL, we need to find the concentration in the original solution. Using the dilution equation:

$C_1 V_1 = C_2 V_2$

where:

• (C_1 = ?)
• (C_2 = 0.0025 , \text{mol L}^{-1})
• (V_1 = 10.00 , \text{mL} = 0.0100 , \text{L})
• (V_2 = 250.0 , \text{mL} = 0.250 , \text{L})

Rearranging gives:

$C_1 = \frac{C_2 V_2}{V_1} = \frac{0.0025 \, \text{mol L}^{-1} \times 0.250 \, \text{L}}{0.0100 \, \text{L}} = 0.0625 \, \text{mol L}^{-1}$

Using the molar mass of nickel, which is approximately 58.69 g/mol, we can calculate the mass:

$\text{mass} = n \times \text{molar mass}$

where:

• (n = 0.0625 , \text{mol L}^{-1} \times 0.0100 , \text{L} = 0.000625 , \text{mol})

Thus:

$\text{mass} = 0.000625 \, \text{mol} \times 58.69 \, \text{g mol}^{-1} \approx 0.0367 \, g$

Based on the options available, the closest mass of the sample of nickel is 0.031 g.