Question

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 be?

Answer 1

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

$m=d\times v=0.050 mL\times 1.00 g/mL=0.050 g$

Moles of water in drops:

$\frac{0.050 g}{18 g/mol}=0.0028 mol$

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

$PV=nRT$ ( ideal gas equation )

$V=\frac{0.0028 mol\times 0.0821 atm L/mol K\times 373 K}{1 atm}$

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.