4.5 g of compound A (M.W. = 90) was used to make 250 mL of its aqueous solution. The molarity of the solution in M is $$x \times 10^{-1}$$. The value of $$x$$ is ______ (Rounded off to the nearest integer)

We are given that 4.5 g of compound A with molecular weight 90 is dissolved to make 250 mL of aqueous solution. We need to find the molarity.

The number of moles of compound A is:

$$n = \frac{\text{mass}}{\text{molar mass}} = \frac{4.5}{90} = 0.05 \text{ mol}$$

The volume of solution is 250 mL, which we convert to litres:

$$V = 250 \text{ mL} = 0.250 \text{ L}$$

Now, molarity is defined as the number of moles of solute per litre of solution:

$$M = \frac{n}{V} = \frac{0.05}{0.250} = 0.2 \text{ M}$$

We are told that the molarity is $$x \times 10^{-1}$$. So:

$$0.2 = x \times 10^{-1}$$ $$x = \frac{0.2}{10^{-1}} = \frac{0.2}{0.1} = 2$$

So, the answer is $$2$$.