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Calculate (a) the actual volume of a molecule of water (b) the radius of a water molecule assuming to be spherical (density of water `= 1 g cm"^(-3)`) |
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Answer» (a) Density of water `= 1g cm^(-3)` Gram molecular mass of water `= 18 g` Gram molecular volume `= ((18g))/((1g cm^(-3)))=18 cm^(3)` `6.022 xx 10^(23)` molecules of water occupy volume `= 18 cm^(3) = 18 c c` 1 molecule of water occupies volume `= (18)/(6.022xx10^(23))=2.99xx10^(-23) c c` (b) Volume of a sphere `= 4//3 pi r^(3) :. 4//3 pi r^(3) = 2.99 xx 10^(-23) c c` or `r^(3)=(2.99xx10^(-23)xx3)/(4xx3.143)" "[becausepi=22//7=3.143]` or `r=((2.99xx10^(-23)xx3)/(4xx3.143))^(1//3)=(7.13xx10^(-24))^(1//3)=1.925xx10^(-8)"cm"=1.925 Ã…`. |
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