How many water molecules in a $0.90*g$ mass of water?

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Answer 1 · 2017-10-25T16:56:28

$3.01xx10^22$ molecules.

The mass of one mole of water is $"18 g"$ . Therefore, $"0.9 g"$ of water is

$"0.9 g"/"18 g/mol" = "0.05 moles"$

of water. One mole of water contains Avagadro's constant, ie, $6.022 xx 10^23$ molecules of water, therefore, $0.05$ moles of water contain

$"0.05 mol" xx (6.022^23"molec.")/("1 mol") = 3.01xx10^22$ $"molecules"$

Answer 2 · 2017-10-25T16:58:41

We find the molar quantity.....and get $"number of water molecules"$ $=3.01xx10^22$ ...

$"Moles of water"=(0.9*g)/(18.01*g*mol^-1)=0.050*mol$

And we know that there are $6.022xx10^23$ molecules in one mole of any substance.....

And so we take the product......

$0.050*molxx6.022xx10^23*mol^-1=3.01xx10^22$ $"water molecules"$ . (And thus the product gives a dimensionless number as required....)

(i) how many oxygen atoms in this molar quantity; and (ii) how many hydrogen atoms......?

Answer 3 · 2017-10-25T17:08:06

$30.1$ x $10^21 mlcs$

$Mass of water(H_2O)$ $= 0.9grams$

$Molar mass of (H_2O)$ $= 1.008$ x $2$ + 1 $6$
$= 18.016gmol^-$

** Number of molecules ** = Mass in grams $/$ Molar mass x $N_A$

$= 0.9/18.016$ x $6.02$ x $10^23$

$= 30.1$ x $10^21$ $mlcs$

How many water molecules in a 0.90*g0.90*g mass of water?