THE VOLUME OF A DROP OF WATER IS 0.0018mL THEN THE NUMBER OF MOLECULES OF WATER PRESENT IN TWO DROP OF WATER AT ROOM TEMPERATURE IS?
Asked by rraj.15266
20th August 2017,
11:25 AM
Answered by Expert
Answer:
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
Answered by Expert
20th August 2017,
12:01 PM
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