Question

4 l, as if the vapor were an ideal gas. Give answer in angstroms, two significant digits. Do not write down units in your answer.Calculate the volumetric size of a water molecule in water vapor at normal conditions, assuming 1 mole of the vapor occupies 22.4 l, as if the vapor were an ideal gas. Give answer in angstroms, two significant digits. Do not write down units in your answer.

Answer 1

Let volumetric size of a water molecule = v

Since, 1 mole of water consists of $6.023\times 10^{23}$ molecules of water.

Thus, total volume of 1 mole of water = $volumetric size \times 6.023\times 10^{23}$

Substitute the value of total volume of 1 mole of water i.e. 22.4 L in above formula.

$22.4 L = v\times 6.023\times 10^{23}$

$v= \frac{22.4 L}{6.023\times 10^{23}}$

Convert the unit litre to angstrom

$v=\frac{22.4 L}{6.023\times 10^{23}}\times \frac{1dm^{3}}{1L}\times \frac{(10^{9})^{3}(A^{o})^{3}}{1 dm^{3}}$

= $3.7\times 10^{-23}\times 10^{27} (A^{o})^{3}$

= $3.7 \times 10^{4}(A^{o})^{3}$

Therefore, the volumetric size of the water molecule is $3.7 \times 10^{4}(A^{o})^{3}$