Calculate the number of molecules present in 34.20 grams of cane sugar C12H22O11

The correct option is 6.022x10^22 (which represents $6.022\times {10}^{22}$ molecules).

Step-by-Step Explanation:

1. Find the molar mass of cane sugar ($C_{12}{H}_{22}{O}_{11}$):
To calculate the molar mass, we sum the atomic masses of all the constituent atoms:
- Carbon (C): $12$ atoms × $12.011\text{g/mol}$ ≈ $144.13\text{g/mol}$
- Hydrogen (H): $22$ atoms × $1.008\text{g/mol}$ ≈ $22.18\text{g/mol}$
- Oxygen (O): $11$ atoms × $15.999\text{g/mol}$ ≈ $175.99\text{g/mol}$
Adding these together:

$\text{Molar mass of}{C}_{12}{H}_{22}{O}_{11}=144.13+22.18+175.99=342.3\text{g/mol}$

For calculation simplicity, we can round the molar mass of cane sugar to $342\text{g/mol}$.

2. Calculate the number of moles in 34.20 grams of cane sugar:
Using the formula:
$\text{Number of moles (n)}=\frac{\text{Given mass}}{\text{Molar mass}}$
Substituting the given values:

$\text{n}=\frac{34.20\text{g}}{342\text{g/mol}}=0.10\text{moles}$

3. Calculate the number of molecules:
One mole of any substance contains Avogadro's number of entities (molecules), which is approximately $6.022\times {10}^{23}$ molecules/mol.
Therefore, the number of molecules in 0.10 moles of cane sugar is:

$\text{Number of molecules}=\text{Number of moles}\times \text{Avogadro's number}$

$\text{Number of molecules}=0.10\times 6.022\times {10}^{23}$

$\text{Number of molecules}=6.022\times {10}^{22}$

Thus, there are exactly $6.022\times {10}^{22}$ molecules present in 34.20 grams of cane sugar.