Question

After 1.00 ml of water is completely vaporized to gas, how many milliliters of vapor are produced at standard temperature and pressure? (Density of water is 1.00 g/mL; molar volume of any gas at STP is 22.4 L/mol.).

Answer 1

Based on your question where 1ml of water is completely vaporized to gas, to get the millimeter or the amount of vapor produced at a standard temperature and pressure you must use a formula and consider the density of water and the molar volume of gas. So the answer is 0.045 ml of water.

Answer 2

Answer : The volume of water vapor will be, 1232 ml

Explanation :

Density is defined as the mass contained per unit volume.

$Density=\frac{Mass}{Volume}$

Given :

Volume of water = 1.00 ml

Density of water = 1.00 g/ml

Putting in the values we get:

$1.00g/ml=\frac{Mass}{1.00ml}$

Mass of water = 1 gram

Now we have tot calculate the moles of water.

$\text{Moles of water}=\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{1g}{18g/mole}=0.055mole$

As, 1 mole of gas occupies 22.4 L of volume at STP

So, 0.055 moles of water vapor will occupy = $0.055\times 22.4=1.232L=1232mL$ of volume      (Conversion factor: 1 L = 1000 mL)

Therefore, the volume of water vapor will be, 1232 ml