Question

Volume occupied by 10^2 molecules of water (density 1 kg/L) iS

Answer 1

Volume occupied by 10^2 molecules of water is $=2.99\times 10^{-21}$ $=2.99\times 10^{-21}$cm³

Explanation:

Molecular weight of water $(H_2O)$ = 18

We know that 1 mole of water contains = $6.023\times 10^{23}$ molecules

Therefore, 100 molecules of water

= $\frac{100}{6.023\times 10^{23}}$

= $\frac{1}{6.023\times 10^{21}}$ moles

= $\frac{18}{6.023\times 10^{21}}$ gram

= $2.99\times 10^{-21}}$ gram

= $2.99\times 10^{-24}}$ kg

Given density of water = 1 kg/L

Therefore, the volume occupied by $2.99\times 10^{-26}}$ kg of water

$=\frac{\text{Mass}}{\text{Density}}$

$=\frac{2.99\times 10^{-24}}{1}$

$=2.99\times 10^{-24}$ Litre

$=2.99\times 10^{-21}$$=2.99\times 10^{-21}$ cm³

Hope this answer is helpful.

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Q: What is the volume occupied by one molecule of water if the density of water is 1gm/ml

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Answer 2

Answer:

2.99 x 10^-24 L

Explanation:

We are given that we have 100 molecules of water.

According to Avagadro principle, one mole of water will contain 6.02x10^23 molecules.

Hence the number of moles in 100 water molecules would be

= 100 / 6.02x10^23

= 1.6 x 10^-22

Molar mass of water = 18 grams

Mass of water = 18 x 1.6 x 10^-22 =2.99 x 10^-21 grams

Density of water is 1 g / cm3.

Hence volume occupied = 2.99 x 10^-21 / 1 =2.99 x 10^-21 cm3

If convert it to Litres = 2.99 x 10^-24 L

Learn more about Avagadro Number;

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