Solveeit Logo

Question

Chemistry Question on States of matter

Volume occupied by one molecule of water (density=1gcm3)(density\, = 1 g\, cm^{-3} ) is

A

9.0×1023cm39.0 \times 10 ^{-23} cm^3

B

6.023×1023cm36.023 \times 10 ^{-23} cm^3

C

3.0×1023cm33.0 \times 10 ^{-23} cm^3

D

5.5×1023cm35.5 \times 10 ^{-23} cm^3

Answer

3.0×1023cm33.0 \times 10 ^{-23} cm^3

Explanation

Solution

(c) 1mole=6.023×1023molecule1 mole= 6.023 \times 10^{23} molecule
18g=6.02×1023molecule18 g = 6.02 \times 10^{23} molecule
18 g = mass of 6.02×10236.02 \times 10^{23} water molecules
Mass of one water molecule =186.023×1023g= \frac {18}{6.023\times 10^{23}} g
Density =1gcm3 = 1 \, g \, cm^{-3}
Volume=MassofonewatermoleculeDensityVolume = \frac {Mass \, of \, one \, water \, molecule}{Density}
=186.023×1023×1=cm3= \frac {18}{6.023 \times 10^{23} \times1 } = cm^3
3.0x1023cm3\simeq 3.0 x 10^{-23} cm^3