Science, asked by Tanisha2909, 1 year ago

Volume occupied by one molecule of water (density
= 1 g cm^-3 ) is

Abhishek474241: The number of moles in 1 litre = 1000 / 18.02 = 55.5 moles. 1 mole of water = 6.022 x 10^23 molecules of water, So 55.5 moles of water = 55.5 x 6.022 x 10^23 = 3.34 x 10^25 molecules of water. The gram molecular mass of water is 18.02. Thus volume occupied by 1 H2O molecule is 2.99 ×10-23 ml.

somnatthh: Wow thanks..... Even i was confused with this one

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Answered by rakhithakur

$6.022 ×10^{23} molecules \\ = 18 g \: \: of water$

$1 molecule= \frac{18}{6.022×10^{23}g} \\ <br />=2.98×10^{-23}$
$<br />Density \: \: \: of \: \: \:water=1 g/ml$

$Volume of 1 molecule of water \\ = \frac{mass \: \: of \: \: 1 \: \: molecule \: \: of \: \: H2O}{density} \\ <br />=2.98×10^-23ml<br />$
hope it help you

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Volume occupied by one molecule of water (density= 1 g cm^-3 ) is ​

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Volume occupied by one molecule of water (density
= 1 g cm^-3 ) is