Which contains more molecules of water, 5.00 cm^3 of liquid water at 0 degrees Celsius or 5.00 cm^3 of water vapor at STP? How many more? What is the ratio of these numbers of molecules in these two samples?
Answers
Answer:
Explanation: You could calculate this:
Molecules of liquid water-
1) find the mass of liquid water using p=m/v (density = mass/volume) --> pv=m.
5.00 x 1 = 5.00 g of water (using the approximate value of the density of water)
2) then using the equation mol = M/Mr (moles = mass/relative molecular mass)
you get 5/18.02 = 0.2774694784 (or 250/901) moles of water
3) using the Avogadro constant you can find the exact number of molecules of water in given moles (You can compare the two just by looking at the number of moles, but you did ask for the number of molecules).
(250/901) x (6.02214086 × 10^23) = 1.671 x 10^23 (4 s.f)
Molecules of water vapor-
I don't know how water can be a gas at standard conditions... but here you go
1) The ideal gas equation is ideal for this calculation (haha). pV = nRT (pressure x volume = number of moles x ideal gas constant x temperature) rearranges to give n=pV/RT. At 101.325 kPa and 273.15 K (STP), moles water vapor in 5.00 cm^3 (5.00x10^-6 m^3) is: (101.325 x 5x10^-6)/(8.3145 x 273.15) 2.230741641 x 10^-7 moles. By now you can see that liquid water has way more moles than water vapor.
2) Again, using the Avogadro constant gives you the exact number of molecules in given moles: (2.230741641 x 10^-7) x (6.02214086 × 10^23) = 1.343 x 10^17
hope this helped