Question

Define Standard solution. Zinc is determined by precipitating and weighing as Zn2Fe(CN)6 . What weight of zinc is contained in a sample that gives 0.348 g precipitate? (Zn=65, Fe=56, C=12, N = 14

Answer 1

[AI-generated answer — Bot GPT-4o]

A standard solution is a solution of known concentration, often used in titration to determine the concentration of an analyte in another solution.

For the second part of your query, let's determine the weight of zinc in the given precipitate of Zn₂Fe(CN)₆.

Firstly, we need to calculate the molar mass of Zn₂Fe(CN)₆:

1. Molar Mass of Zn₂Fe(CN)₆:
- Zn: \( 65 \text{ g/mol} \times 2 = 130 \text{ g/mol} \)
- Fe: \( 56 \text{ g/mol} \times 1 = 56 \text{ g/mol} \)
- C: \( 12 \text{ g/mol} \times 6 = 72 \text{ g/mol} \)
- N: \( 14 \text{ g/mol} \times 6 = 84 \text{ g/mol} \)

Adding these together:
\[ 130 + 56 + 72 + 84 = 342 \text{ g/mol} \]

Now, we have 0.348 g of Zn₂Fe(CN)₆ precipitate. The molar mass of Zn₂Fe(CN)₆ is 342 g/mol, so we can find the number of moles of Zn₂Fe(CN)₆:

\[ \text{Moles of Zn}_{2}\text{Fe(CN)}_6 = \frac{0.348 \text{ g}}{342 \text{ g/mol}} \approx 0.0010175 \text{ moles} \]

In one mole of Zn₂Fe(CN)₆, there are 2 moles of zinc (Zn). Therefore, the moles of zinc in the sample are:

\[ 2 \times 0.0010175 \text{ moles} \approx 0.002035 \text{ moles} \]

Finally, to find the mass of zinc:

\[ \text{Mass of Zn} = 0.002035 \text{ moles} \times 65 \text{ g/mol} \approx 0.132275 \text{ g} \]

So, the weight of zinc contained in the sample that gives 0.348 g precipitate of Zn₂Fe(CN)₆ is approximately 0.132 g.