The number of atoms in 8 g of sodium is $$x \times 10^{23}$$. The value of x is _________. (Nearest integer)
Given: N$$_A$$ = $$6.02 \times 10^{23}$$ mol$$^{-1}$$, Atomic mass of Na = 23.0 u

We know that the number of atoms in any given sample can be found with the basic mole concept formula:

$$\text{Number of atoms} \;=\; \text{Number of moles}\;\times\;N_A$$

First, we must evaluate the number of moles present in the 8 g sample of sodium. We use the definition of a mole, which states:

$$\text{Number of moles} \;=\; \frac{\text{Given mass (in g)}}{\text{Molar mass (in g mol}^{-1})}$$

For sodium (Na) the given data are:

Given mass = $$8 \text{ g}$$,   Molar (atomic) mass = $$23.0 \text{ g mol}^{-1}$$.

Substituting these values we get:

$$\text{Number of moles of Na} \;=\; \frac{8}{23.0}$$

Now we multiply this mole value by Avogadro’s constant to obtain the number of atoms:

$$\text{Number of atoms} \;=\; \left(\frac{8}{23.0}\right)\times (6.02\times10^{23})$$

Carrying out the multiplication in the numerator first:

$$8 \times 6.02 = 48.16$$

So we have:

$$\text{Number of atoms} \;=\; \frac{48.16}{23.0}\times10^{23}$$

Next we perform the division $$48.16 \div 23.0$$:

$$\frac{48.16}{23.0} \approx 2.093$$

Thus,

$$\text{Number of atoms} \;\approx\; 2.093\times10^{23}$$

The question states that this number of atoms can be written as $$x\times10^{23}$$, so by comparison we have $$x \approx 2.093$$. Rounding to the nearest integer gives:

$$x = 2$$

So, the answer is $$2$$.