Question

The density of water at 4°C is 1.0 X 10³kg m-3 The volume occupied by one molecule of water is
approximately (NA = 6.0 x 1023)
(A) 3.0 x 10-23 mL
(B) 6.0 x 10-22 mL
(D) 9.0 x 10-23 mL
(C)3.0 x 10-21 mL​

Answer 1

Answer:

Option A is correct.

Explanation:

We know that volume of

${10}^{3} kg \: of \: water \: = 1 {m}^{3} \\ = > volume \: of \: 1kg \: water = \frac{1}{1000} {m}^{3} = (\frac{1}{ {10}} )^{3} \\ or \: = \: one \: decimeter \: cube$

and we also know that 1dm^3 = one litre

Hence the mass of one kg water = 1 litre

and one gram = 1 mL

We also know that one mole of water = 18 gram = 18mL contain 6.0 x 1023 molecules

hence the volume of one molecule of water

$= \frac{18}{6 \times {10}^{23} } \: ml = 3 \times {10}^{ - 23} ml$

Answer 2

Answer:

A

Explanation:

Given density =1×103 kgm−3=1 gm−3

Molar mass of water =18 gm/mole

No. of moles in 1 gm =1/18=0.55 moles

18 gm contain molecules of H2O = 6.022×1023

1 gm contain molecules of H2O = 186.022×1023=0.33×1023

Volume occupied by 1 molecule of water is =0.33×10231=2.99×10−23