10.0 mL of 0.05 M KMnO$$_4$$ solution was consumed in a titration with 10.0 mL of given oxalic acid dihydrate solution. The strength of given oxalic acid solution is _________ $$\times 10^{-2}$$ g/L. (Round off to the nearest integer)

First, we note that in an acidic medium the permanganate ion $$\text{MnO}_4^-$$ is reduced from the oxidation state $$+7$$ to $$+2$$. According to the change in oxidation number, each mole of $$\text{KMnO}_4$$ therefore gives up $$5$$ electrons, so its n-factor is $$5$$ in such titrations.

The molarity of the potassium permanganate solution is given as $$0.05\ \text{M}$$ and its volume used is $$10.0\ \text{mL}=0.010\ \text{L}$$. We convert this molarity into normality by multiplying by the n-factor:

$$N_1 = M_1 \times n\text{-factor} = 0.05 \times 5 = 0.25\ \text{N}.$$

The number of equivalents of $$\text{KMnO}_4$$ actually used is then

$$\text{Equivalents of }\text{KMnO}_4 = N_1 \times V_1 = 0.25\ \text{N} \times 0.010\ \text{L} = 0.0025\ \text{eq}.$$

During titration these equivalents must exactly equal the equivalents furnished by the oxalic acid dihydrate solution, because in any redox titration the two reactants exchange the same total number of electrons.

Hence for $$10.0\ \text{mL}=0.010\ \text{L}$$ of oxalic acid dihydrate solution we must also have

$$\text{Equivalents of }(\text{COOH})_2\cdot 2\text{H}_2\text{O}=0.0025\ \text{eq}.$$

Oxalic acid (or the oxalate ion that forms in acid medium) is oxidised from carbon oxidation state $$+3$$ to $$+4$$ in carbon dioxide, a change of $$+1$$ per carbon atom and hence $$+2$$ per molecule. Thus each mole of oxalic acid loses $$2$$ electrons, giving it an n-factor of $$2$$ in this reaction.

Its normality therefore is

$$N_2=\frac{\text{equivalents}}{\text{volume in L}}=\frac{0.0025}{0.010}=0.25\ \text{N}.$$

We convert this normality into molarity by dividing by the n-factor:

$$M_2=\frac{N_2}{n\text{-factor}}=\frac{0.25}{2}=0.125\ \text{M}.$$

Now we need the strength, i.e. the mass of oxalic acid dihydrate per litre. The molar mass is calculated as follows:

$$\text{Molar mass of }H_2C_2O_4 = 2(1)+2(12)+4(16)=90\ \text{g mol}^{-1},$$

and with two waters of crystallisation

$$2\text{H}_2\text{O}=2\bigl[2(1)+16\bigr]=36\ \text{g mol}^{-1}.$$

So

$$M_{\text{dihydrate}} = 90+36 = 126\ \text{g mol}^{-1}.$$

Strength (grams per litre) is

$$S = M_2 \times M_{\text{dihydrate}} = 0.125\ \text{mol L}^{-1}\times126\ \text{g mol}^{-1} = 15.75\ \text{g L}^{-1}.$$

To fit the required form $$\text{number}\times10^{-2}\ \text{g L}^{-1}$$ we write

$$15.75\ \text{g L}^{-1}=1575\times10^{-2}\ \text{g L}^{-1}.$$

Rounded to the nearest integer, the number is $$1575$$.

So, the answer is $$1575$$.