Question

Volume occupied by 10^2 molecules of water (density: 1 kg/L) is: a)2.99×10^-20 cm3 b)5.61×10^−21 cm3 c)5.61×10^−20 cm3 d)2.99×10^−21 cm3

Answer 1

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Explanation:

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Answer 2

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

Therefore option (d) is correct.

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

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

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

Hope this answer is helpful.

Know More:

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

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