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.
I wanna say 3,398.08 but let me know if that sounds very off
The answer is 6.022• 10^23 atoms
Answer:
Adding a solution containing an anion that forms an insoluble salt with only one of the metal ions.
Explanation:
The student have in solution Ag⁺ and Cu²⁺ ions but he just want to analyze the silver, that means he need to separate ions.
Centrifuging the solution to isolate the heavier ions <em>FALSE </em>Centrifugation allows the separation of a suspension but Ag⁺ and Cu²⁺ are both soluble in water.
Adding enough base solution to bring the pH up to 7.0 <em>FALSE </em>At pH = 7,0 these ions are soluble in water and its separation will not be possible.
Adding a solution containing an anion that forms an insoluble salt with only one of the metal ions <em>TRUE </em>For example, the addition of Cl⁻ will precipitate the Ag⁺ as AgCl(s) allowing its separation.
Evaporating the solution to recover the dissolved nitrates. <em>FALSE</em> . Thus, you will obtain the nitrates of these ions but will be mixed doing impossible its separation.
I hope it helps!