Question

water is a unique material in that the density of the solid is lower than the density of the liquid (which is why ice forms at the top of a pond and why ice floats in our drinks). if the density for ice at 0C is .917g/mL and the density for water at 0C is .999 g/mL, what is the calculated free space (as %) for each of these materials. you will need to estimate the volume of water as the sum of 2 H atoms and 1 O atom with radii 37 and 66 pm respectively. note that you will also have to assume a quantity of water to perform this exercise

Answer 1

% Free space in water =
`(9.945* 10^(-7) )/(1*10^(-6) )`×100 = 99.45%

% Free space in ice =
`(9.98* 10^(-7) )/(1*10^(-6) )`×100 = 99.8%

Step-by-step explanation:

As given ,

Density for ice at 0⁰C = 0.917 g/ml

Density for water at 0⁰C = 0.999 g/ml

Radii of H atoms = 37 pm

Radii of O atoms = 66 pm

Now,

Consider 1 ml of water = 1 cm²

As , we know that mass of water in 1 cm² = 0.999 g

Moles of water =
`(0.999)/(18) = 0.056`

Volume of H₂O = 1.624×
`10^(-31)` m²

Now,

Volume occupied by water = 0.056×6.022×
`10^(23)`× 1.624×
`10^(-31)` m²

= 5.48×
`10^(-9)` m²

⇒Volume occupied by water = 5.48×
`10^(-9)` m²

Now,

Free space = 1×
`10^(-6)` - 5.48×
`10^(-9)` = 9.95×
`10^(-7)` m²

% Free space =
`(9.945* 10^(-7) )/(1*10^(-6) )`×100 = 99.45%

Now,

Consider 1 ml of ice = 1 cm²

S.I unit of ice = 1×
`10^(-6)` m²

As , we know that mass of water in 1×
`10^(-6)` m² = 0.917 g

Moles of ice =
`(0.917)/(18) = 0.012`

Volume of H₂O = 6.022×
`10^(23)` ×0.012

Volume of ice unit =
`(4)/(3) \pi (37*10^(-12))^(3) *2 + (4)/(3) \pi (66*10^(-12))^(3) = 1.624*10^(-31)m^(3)`

Now,

Volume occupied by water = 0.012×6.022×
`10^(23)`× 1.624×
`10^(-31)` m²

= 1.17×
`10^(-9)` m²

⇒Volume occupied by water = 1.17×
`10^(-9)` m²

Now,

Free space = 1×
`10^(-6)` - 1.17×
`10^(-9)` = 9.98×
`10^(-7)` m²

% Free space =
`(9.98* 10^(-7) )/(1*10^(-6) )`×100 = 99.8%