824 g NH3 (1 mol/17 g NH3) (4 mol NO/4mol NH3)
48.47 moles NO
Answer:
Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water <em><u>to the same extent</u></em> by adding 1 mol of
to 1 kg of water.
Explanation:
1) Moles of NaCl ,
Mass of water = m= 1 kg = 1000 g
Moles of water = 
Vapor pressure of the solution = 
Vapor pressure of the pure solvent that is water = 
Mole fraction of solute(NaCl)= 



The vapor pressure for the NaCl solution at 17.19 Torr.
2) Moles of sucrose ,
Mass of water = m = 1 kg = 1000 g
Moles of water = 
Vapor pressure of the solution = 
Vapor pressure of the pure solvent that is water = 
Mole fraction of solute ( glucose)= 



The vapor pressure for the glucose solution at 17.19 Torr.
p = p' = 17.19 Torr
Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent by adding 1 mol of
to 1 kg of water.
When a solvent has as much of the dilute dissolved in it as possible, then it is saturated.
If you were to heat the water, its capacity would increase and would then be super-saturated because it has more dissolved in it than possible as room temp.
Since there is no heating being done, the water is just saturated.
Hope that helps!