1) A solution contains 27.79 grams of NaCl in .909 kg water at 25 degrees celciu
ID: 703965 • Letter: 1
Question
1) A solution contains 27.79 grams of NaCl in .909 kg water at 25 degrees celcius. What is the vapor pressure of the solution? (The molar mass of NaCl is 58.44 g/mol. The molar mass of water is 18.02 g/mol.) and (The vapor pressure of water is 23.7 torr at 25 degrees Celcius
2)The freezing point of benzene is 5.5 degrees celcius. What is the molar mass of a nonionizing solute if 1.80 grams of the solute added to 24.81g of benzene lowers the freezing point to 3.22. The Kf of benzene is -5.12 degrees C/m.
Explanation / Answer
Ans 1
Moles of NaCl = mass/molecular weight
= 27.79g / 58.44g/mol
= 0.4755 mol
Moles of water = 0.909*1000 g / 18.02g/mol
= 50.444 mol
Total Moles = 50.444 + 0.4755 = 50.9195 mol
Mol fraction of water = moles of water / total moles
= 50.444/50.9195 = 0.99
Vapor pressure of solution = mol fraction of water x vapor pressure of water
= 0.99 x 23.7
= 23.463 torr
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