Respuesta :
Explanation:
The given data is as follows.
P = 28.3 mm = [tex]\frac{28.3}{760}[/tex], V = 600 mL = [tex]600 ml \frac{0.001 L}{1 ml}[/tex] = 0.6 L
R = 0.082 L atm/mol K, T = (28 + 273) K = 301 K
Therefore, according to ideal gas law PV = nRT. Hence, putting the values into this equation calculate the number of moles as follows.
PV = nRT
[tex]\frac{28.3}{760} \times 0.6 L[/tex] = [tex]n \times 0.082 L atm/mol K \times 301 K[/tex]
n = [tex]9.03 \times 10^{-4}[/tex] mol
As it is known that number of moles equal to mass divided by molar mass. Hence, mass of water vapor present will be calculated as follows. (molar mass of water is 18 g/mol)
No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
[tex]9.03 \times 10^{-4}[/tex] mol = [tex]\frac{mass}{18 g/mol}[/tex]
mass = [tex]162.5 \times 10^{-4}g[/tex]
= [tex]163 \times 10^{-4}g[/tex] (approx)
Since, 1 g = 1000 mg. Therefore, [tex]163 \times 10^{-4}g[/tex] will be equal to [tex]163 \times 10^{-4}g \times \frac{10^{3}mg}{1 g}[/tex]
= 16.3 mg
Thus, we can conclude that mass of water vapor present is 16.3 mg.
