When 10 mL of an aqueous solution of Fe$$^{2+}$$ ions was titrated in the presence of dil H$$_2$$SO$$_4$$ using diphenylamine indicator, 15 mL of 0.02 M solution of K$$_2$$Cr$$_2$$O$$_7$$ was required to get the end point. The molarity of the solution containing Fe$$^{2+}$$ ions is $$x \times 10^{-2}$$ M. The value of x is ___. (Nearest integer)

We are told that 10 mL of a solution containing Fe$$^{2+}$$ ions is titrated with 0.02 M K$$_2$$Cr$$_2$$O$$_7$$ in the presence of dilute H$$_2$$SO$$_4$$, using diphenylamine as the indicator. The titration requires 15 mL of the dichromate solution to reach the end point. Our task is to find the molarity of the Fe$$^{2+}$$ solution and express it in the form $$x \times 10^{-2}\,{\rm M}$$, then determine the integer $$x$$.

First, we recall the balanced redox reaction that occurs in acidic medium between dichromate ions and ferrous ions:

$$\text{Cr}_2\text{O}_7^{2-} + 14\,\text{H}^+ + 6\,\text{Fe}^{2+} \;\longrightarrow\; 2\,\text{Cr}^{3+} + 6\,\text{Fe}^{3+} + 7\,\text{H}_2\text{O}$$

From this equation we see clearly that 1 mole of dichromate reacts with 6 moles of Fe$$^{2+}$$.

Now we proceed step by step, starting with the dichromate solution:

Using the definition of molarity, the number of moles is given by

$$n = M \times V$$

where $$M$$ is molarity in mol L$$^{-1}$$ and $$V$$ is volume in litres.

The dichromate data are:

$$M_{\text{Cr}_2\text{O}_7^{2-}} = 0.02\,{\rm M}, \qquad V_{\text{Cr}_2\text{O}_7^{2-}} = 15\,\text{mL} = 0.015\,\text{L}$$

Substituting, we obtain

$$n_{\text{Cr}_2\text{O}_7^{2-}} = 0.02 \times 0.015 = 0.00030\;\text{mole}$$

According to the stoichiometric ratio 1 : 6, the moles of Fe$$^{2+}$$ oxidised are

$$n_{\text{Fe}^{2+}} = 6 \times n_{\text{Cr}_2\text{O}_7^{2-}} = 6 \times 0.00030 = 0.00180\;\text{mole}$$

The volume of the Fe$$^{2+}$$ solution taken is

$$V_{\text{Fe}^{2+}} = 10\,\text{mL} = 0.010\,\text{L}$$

Again invoking the molarity formula,

$$M_{\text{Fe}^{2+}} = \dfrac{n_{\text{Fe}^{2+}}}{V_{\text{Fe}^{2+}}} = \dfrac{0.00180}{0.010} = 0.18\;\text{M}$$

To express this in the required format, we write

$$0.18\;\text{M} = 18 \times 10^{-2}\;\text{M}$$

Thus, $$x = 18$$.

So, the answer is $$18$$.