100 mL of Na$$_3$$PO$$_4$$ solution contains 3.45 g of sodium. The molarity of the solution is _________ $$\times 10^{-2}$$ mol L$$^{-1}$$. (Nearest integer)
[Atomic Masses - Na : 23.0u, O : 16.0u, P : 31.0u]

We are told that in $$100 \text{ mL}$$ of the solution of $$\text{Na}_3\text{PO}_4$$, the total mass of sodium present is $$3.45 \text{ g}$$. Our task is to convert this information into the molarity of $$\text{Na}_3\text{PO}_4$$.

First we convert the given volume to litres, because molarity requires litres:

$$100 \text{ mL}= \frac{100}{1000} \text{ L}=0.10 \text{ L}$$

Next, we calculate the moles of sodium atoms present. We recall the definition:

Number of moles $$n=\dfrac{\text{Given mass}}{\text{Molar mass}}$$.

For sodium, the atomic mass is given as $$23.0 \text{ u}$$, so

$$n(\text{Na})=\dfrac{3.45 \text{ g}}{23.0 \text{ g mol}^{-1}}=0.150 \text{ mol}$$.

Now we link the sodium atoms to the formula units of $$\text{Na}_3\text{PO}_4$$. Each formula unit has three sodium atoms:

$$\text{Na}_3\text{PO}_4 \longrightarrow 3\;\text{Na atoms per formula unit}$$.

Therefore, the moles of $$\text{Na}_3\text{PO}_4$$ present are one-third of the moles of sodium atoms:

$$n(\text{Na}_3\text{PO}_4)=\dfrac{0.150 \text{ mol}}{3}=0.050 \text{ mol}$$.

With the moles of solute and the volume of solution, we apply the formula for molarity $$M$$:

$$M=\dfrac{\text{moles of solute}}{\text{volume of solution in L}}$$.

Substituting the numbers,

$$M=\dfrac{0.050 \text{ mol}}{0.10 \text{ L}}=0.50 \text{ mol L}^{-1}$$.

To match the required format $$\times 10^{-2} \text{ mol L}^{-1}$$, we rewrite $$0.50$$ as $$50 \times 10^{-2}$$:

$$0.50 \text{ mol L}^{-1}=50 \times 10^{-2} \text{ mol L}^{-1}$$.

The nearest integer to $$50$$ is clearly $$50$$.

Hence, the correct answer is Option 50.