The molarity of the solution prepared by dissolving 6.3 g of oxalic acid (H$$_2$$C$$_2$$O$$_4$$.2H$$_2$$O) in 250 mL of water in mol L$$^{-1}$$ is $$x \times 10^{-2}$$. The value of x is _________. (Nearest integer)
[Atomic mass : H : 1.0, C : 12.0, O : 16.0 J]

We have to find the molarity of the solution prepared by dissolving 6.3 g of oxalic acid dihydrate, whose formula is $$\text{H}_2\text{C}_2\text{O}_4\cdot 2\text{H}_2\text{O}$$.

First, we calculate its molar mass. The formula tells us how many atoms of each element are present. We now multiply the number of atoms by their respective atomic masses and then sum all the contributions.

For the anhydrous part $$\text{H}_2\text{C}_2\text{O}_4$$:

$$\text{Mass of }2\text{H}=2\times 1 = 2\;\text{g}$$ $$\text{Mass of }2\text{C}=2\times 12 = 24\;\text{g}$$ $$\text{Mass of }4\text{O}=4\times 16 = 64\;\text{g}$$

Adding them, we obtain

$$90\;\text{g mol}^{-1}$$

Now we must add the contribution from the two water molecules of crystallisation, $$2\text{H}_2\text{O}$$.

For one water molecule $$\text{H}_2\text{O}$$ the molar mass is

$$2\times 1 + 16 = 18\;\text{g mol}^{-1}$$

So for two water molecules we have

$$2 \times 18 = 36\;\text{g mol}^{-1}$$

Therefore, the complete molar mass of $$\text{H}_2\text{C}_2\text{O}_4\cdot 2\text{H}_2\text{O}$$ is

$$90 + 36 = 126\;\text{g mol}^{-1}$$

Now we calculate the number of moles of oxalic acid actually present in 6.3 g. The defining formula for moles is

$$\text{Moles} = \dfrac{\text{Given mass}}{\text{Molar mass}}$$

Substituting the values,

$$\text{Moles} = \dfrac{6.3}{126} = 0.05\;\text{mol}$$

The given solution volume is 250 mL. We must convert this into litres because molarity uses units of litres:

$$250\;\text{mL} = 250 \times 10^{-3}\;\text{L} = 0.25\;\text{L}$$

By definition, molarity $$M$$ is

$$M = \dfrac{\text{Moles of solute}}{\text{Volume of solution in litres}}$$

Substituting the numbers,

$$M = \dfrac{0.05}{0.25} = 0.20\;\text{mol L}^{-1}$$

We are asked to express the answer in the form $$x \times 10^{-2}$$. To write 0.20 in that format, notice that

$$0.20 = 20 \times 10^{-2}$$

Thus,

$$x = 20$$

So, the answer is $$20$$.