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

Answer:

% Free space in water = $\frac{9.945* 10^{-7} }{1*10^{-6} }$×100 = 99.45%

% Free space in ice  = $\frac{9.98* 10^{-7} }{1*10^{-6} }$×100 = 99.8%

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 = $\frac{0.999}{18} = 0.056$

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 = $\frac{9.945* 10^{-7} }{1*10^{-6} }$×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 = $\frac{0.917}{18} = 0.012$

Volume of H₂O = 6.022×$10^{23}$ ×0.012

Volume of ice unit = $\frac{4}{3} \pi (37*10^{-12})^{3} *2 + \frac{4}{3} \pi (66*10^{-12})^{3} = 1.624*10^{-31}m^{3}$

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 = $\frac{9.98* 10^{-7} }{1*10^{-6} }$×100 = 99.8%