Which contains more molecules of water: 5.00 #cm^3# of ice at 0°C or 5.00 #cm^3# of liquid water at 0.°C? How many more? What is the ratio of the numbers of molecules in these two samples?

So we will have to get these data...shortly. You add an ice cube to a glass of water...the ice FLOATS.....you sail your luxury liner thru the North Atlantic, you do run the risk of ramming an ice-berg. And so clearly, LIQUID WATER is DENSER than SOLID ICE under normal conditions, and this is on what the question depends... This is an UNUSUAL property, which is why it is emphasized in your syllabus...to illustrate the unusual properties of water...

A #5.00*cm^3# volume of water has a mass of #5.00*cm^3xxunderbrace(1.00*g*cm^-3)_"density of water"=5.00*g#...and a number of water molecules given by the Avocado number...#N_A=6.022xx10^23*mol^-1#...

#n_"number of water molecules in water"=(5.00*g)/(18.01*g*mol^-1)xxN_A=1.67xx10^23*"water molecules"#

This site quotes the density of ice as...#rho_"ice"=0.917*g*cm^-3#..

#n_"number of water molecules in ice"=(5.00*cm^3xx0.917*g*cm^-3)/(18.01*g*mol^-1)xxN_A=1.53xx10^23*"water molecules"#

You can do the ratio...

First we will use density to determine the mass of each volume. Then we can use mass and molar mass to determine the number of moles. Then we can calculate the number of molecules by multiplying the moles by #6.022xx10^23# #"molecules/mol"#.

WATER ICE

The density of water ice at #0^@"C"# is #"0.9167 g/cm"^3"#. We can use the density formula to determine the mass of #"5.00 cm"^3"# of ice.

#D=M/V#,

where #D# is density, #M# is mass, and #V# is volume.

To get mass, multiply #VxxD#.

#M_("H"_2"O")=5.00color(red)cancel(color(black)("cm"^3))xx(0.9167"g")/(1color(red)cancel(color(black)("cm"^3)))="4.5835 g"#

To get moles of ice, divide the mass of liquid water by the molar mass of water #("18.015 g/mol")#. Do this by multiplying the mass by the inverse of the molar mass (mol/g).

#4.5835color(red)cancel(color(black)("g H"_2"O"))xx(1"mol H"_2"O")/(18.015color(red)cancel(color(black)("g H"_2"O")))="0.2544 mol H"_2"O"#

To calculate the number of molecules of ice, multiply mol #"H"_2"O"# by #6.022xx10^23# #"molecules/mol"#.

#0.2544color(red)cancel(color(black)("mol")) "H"_2"O"xx(6.022xx10^23"molecules")/(1color(red)cancel(color(black)("mol")))=1.53xx10^23# #"molecules H"_2"O"#

LIQUID WATER

The density of liquid water at #0^@"C"# is #"0.9998 g/cm"^3"#. Again, we can use the density formula to determine the mass of #"5.00 cm"^3"# of ice by multiplying the volume by the density.

#M_("H"_2"O")=5.00color(red)cancel(color(black)("cm"^3))xx(0.9998"g")/(1color(red)cancel(color(black)("cm"^3)))="4.999 g"#

To get moles of water, divide the mass of liquid water by the molar mass of water #("18.015 g/mol")#. Do this by multiplying the mass by the inverse of the molar mass (mol/g).

#4.999color(red)cancel(color(black)("g H"_2"O"))xx(1"mol H"_2"O")/(18.015color(red)cancel(color(black)("g H"_2"O")))="0.2775 mol H"_2"O"#

To calculate the number of molecules of liquid water, multiply mol #"H"_2"O"# by #6.022xx10^23# #"molecules/mol"#.

#0.2775color(red)cancel(color(black)("mol H"_2"O"))xx(6.022xx10^23"molecules")/(1color(red)cancel(color(black)("mol")))=1.67xx10^23# #"molecules H"_2"O"#

DIFFERENCE IN NUMBER OF MOLECULES

#1.67xx10^23-1.53xx10^23=1.4xx10^22# #"molecules"#

RATIO

We can just divide the coefficients since the exponents are the same.

Ratio of molecules of liquid water to water ice is:

#"1.67 liquid water"/"1.53 water ice"="1.09: 1#