Question

Problem 2. Calculate the number of water
molecules in a drop of water whose volume is 0.045
ml at 277 K and 1 atm. pressure.​

Answer 1

Explanation:

At RT i.e. 25 C, the density of water is 1 g/L.

Mass of water = Density x volume = 1 g/L x 0.0018 x 10-3 L = 1.8 x 10-3 g

18 g water = 1 mole

1mole water contains 6.023 x 1023 molecules of water

So, 18 g water cotains 6.023 x 1023 molecules of water

Then, 1.8 x 10-3 g water contains 1.8 x 10-3 x 6.023 x 1023 / 18 g = 6.023 x 1019 molecules of water

Answer 2

Explanation:

$\huge\bf{\mid{\overline{\underline{ANSWER:-}\mid}}}$

$\huge\bold{Given:-} \\ For \: \: water \: \: drop, \\ \\ V = 0.045 mL = 4.5 \times {10}^{-5} L \\ \\ \huge\bold{Formula:-} \\ \\ PV = nRT \\ n = \frac{w}{M} \\ \\ PV = nRT \\ \\ PV = \frac{w}{M} \times R \times T \\ \\ w = \frac{PVM}{RT} \\ \\ w = \frac{1 \times 18 \times 4.5 \times {10}^{-5}}{0.082 \times 277} \\ \\ w = 3.5 \times {10}^{-5} g \\ \\ n = \frac{w}{M} = 2 \times {10}^{-5} \\ \\ No. \: of \: Molecules = n \times N_{A} \\ \\ No. \: of \: Molecules = 2 \times {10}^{-5} \times 6.022 \times {10}^{23} \\ \\ \huge\boxed{ No. \: of \: Molecules = 12.44 \times {10}^{18} }$