Question

0.35 g FeSO4 is dissolved to give 100 mL stock solution. 25 mL of this stock solution in turn is diluted to 250 mL. Calculate the [Fe2+] (mol/L) in the final solution.

Answer 1

Answer on the question #48867, Chemistry, Other

Question:

0.35 g FeSO4 is dissolved to give 100 mL stock solution. 25 mL of this stock solution in turn is diluted to 250 mL. Calculate the [Fe2+] (mol/L) in the final solution.

Solution:

Let's calculate the concentration of the stock solution:

$$c \left(\text {stock}\right) = \frac {n \left(F e S O _ {4}\right)}{V (\text {st.sol.})} = \frac {m \left(F e S O _ {4}\right)}{M \left(F e S O _ {4}\right) V (\text {st.sol.})} = \frac {0 . 3 5}{1 5 1 , 9 0 8 * 0 . 1} = 0. 0 2 3 \frac {\text {mol}}{L}$$

Then the dilution of stock will produce:

$$c \left(F e ^ {2 +}\right) = \frac {c (\text {stock}) V (\text {stock})}{V} = \frac {0 . 0 2 3 * 2 5}{2 5 0} = 2. 3 * 1 0 ^ {- 3} \frac {\text {mol}}{L}$$

Answer: $2.3 * 10^{-3} \frac{mol}{L}$

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