Respuesta :
The question is incomplete and complete question is :
It was important that the flask be completely dry before the unknown liquid was added so that water present would not vaporize when the flask was heated. A typical single drop of liquid water has a volume of approximately 0.050 mL. Assuming the density of liquid water is 1.00 g/mL, how many moles of water are in one drop of liquid, and what volume would this amount of water occupy when vaporized at 100°C and 1atm ?
Answer:
0.0028 moles of water are in one drop of liquid.
Volume of 0.0028 moles of water occupy when vaporized at 100°C and 1 atm is 86 mL.
Explanation:
Volume of of drop = v = 0.050 mL
Mass of drop = m
Density of water = d = 1.00 g/mL
[tex]m=d\times v=0.050 mL\times 1.00 g/mL=0.050 g[/tex]
Moles of water in drops:
[tex]\frac{0.050 g}{18 g/mol}=0.0028 mol[/tex]
0.0028 moles of water are in one drop of liquid.
Pressure at which 0.0028 moles are vaporized = P = 1 atm
Temperature at which 0.0028 moles are vaporized = T = 100°C = 100+273 K = 373 K atm
Volume of moles of water = V
Moles of water = n = 0.0028 mol
[tex]PV=nRT[/tex] ( ideal gas equation )
[tex]V=\frac{0.0028 mol\times 0.0821 atm L/mol K\times 373 K}{1 atm}[/tex]
V = 0.086 L = 86 mL ( 1L = 1000 mL)
Volume of 0.0028 moles of water occupy when vaporized at 100°C and 1 atm is 86 mL.
