To find the weight of $$1 \times 10^{22}$$ molecules of $$\text{CuSO}_4.5\text{H}_2\text{O}$$ (Copper(II) sulfate pentahydrate), we will follow several steps, including understanding the molar mass of the compound, Avogadro's number, and how to use these to find the mass of the given number of molecules.

First, let's calculate the molar mass of $$\text{CuSO}_4.5\text{H}_2\text{O}$$:

• Copper (Cu) = $$63.55 \, \text{g/mol}$$


• Sulfur (S) = $$32.07 \, \text{g/mol}$$


• Oxygen (O) = $$16.00 \, \text{g/mol}$$


• Hydrogen (H) = $$1.008 \, \text{g/mol}$$

The molar mass of $$\text{CuSO}_4.5\text{H}_2\text{O}$$ is calculated as follows:

$M = (\text{Cu}) + (\text{S}) + 4(\text{O}) + 5[(2(\text{H})) + (\text{O})] = 63.55 + 32.07 + 4(16.00) + 5[2(1.008) + 16.00] = 63.55 + 32.07 + 64.00 + 5(2.016 + 16.00) = 63.55 + 32.07 + 64.00 + 5(18.016) = 159.62 + 90.08 = 249.7 \, \text{g/mol}$

Now, knowing that Avogadro's number ($$N_A$$) is $$6.022 \times 10^{23}$$ molecules/mol, which represents the number of molecules in one mole of any substance, we can find the mass of $$1 \times 10^{22}$$ molecules of $$\text{CuSO}_4.5\text{H}_2\text{O}$$ using the proportion:

$\frac{1 \times 10^{22} \, \text{molecules}}{x \, \text{g}} = \frac{6.022 \times 10^{23} \, \text{molecules}}{249.7 \, \text{g}}$

Solving for $$x$$ gives us the mass of the $$1 \times 10^{22}$$ molecules:

$x = \frac{1 \times 10^{22} \, \text{molecules} \times 249.7 \, \text{g}}{6.022 \times 10^{23} \, \text{molecules}} = \frac{249.7}{6.022} \times 10^{-1} \approx 41.45 \times 10^{-1} \, \text{g} = 4.145 \, \text{g}$

Therefore, the weight of $$1 \times 10^{22}$$ molecules of $$\text{CuSO}_4.5\text{H}_2\text{O}$$ is approximately $$4.145 \, \text{g}$$.