Please show steps. Thank you. A solution is prepared by mixing 75 g of methanol
ID: 1016147 • Letter: P
Question
Please show steps. Thank you.
A solution is prepared by mixing 75 g of methanol (CH3OH, 32.04 g/mol) with 25 g of ethanol (CH3CH2OH, 46.07 g/mol). What is the partial pressure of methanol in the vapor phase at 20°C?
Substance
Vapor Pressure
at 20°C (torr)
Methanol
92
Ethanol
45
69 torr
57 torr
80 torr
73 torr
83 torr
The vapor pressure of benzene (C6H6, 78.12 g/mol) at 25°C is 0.1252 atm. What is the change in the vapor pressure when 10.00 g naphthalene (C10H8, 128.2 g/mol) is dissolved in 0.2000 kg C6H6?
-0.9704 atm
-0.0296 atm
-0.0036 atm
-0.1215 atm
-0.9704 atm
At 25°C, the vapor pressure of pure water is 25.756 mmHg. Starting with 250.0 g water and solid glucose (C6H12O6, 180.2 g/mol), you must create an aqueous solution that has a vapor pressure reduction of 2.000 mmHg. How many grams of glucose do you need?
1.17 g
14.1 g
19.5 g
166 g
211 g
Explanation / Answer
a. Number of moles of CH3OH n1= 75/32.04 = 2.34 mol
Number of moles of CH3CH2OH n2= 25/46.07 = 0.54 mol
Total pressure P = (n1 P1 + n2 P2)/(n1 + n2) = (2.34 * 92 + 0.54 * 45)/(2.34+.54) = 83.1875 torr
Hence, partial pressure of CH3OH = (n1/n1+n2) * P = (2.34/2.34+0.54) * 83.1875 torr = 68.48 torr ~69 torr
b. Moles of C6H6 n1= 200/78.12 mol= 2.56 mol
moles of naphthalene n2 = 10/128.2 mol = 0.078 mol
mole fraction of C6H6 in the solution = 2.56/(2.56+0.078) = 0.9704
Hence, the changed vapour pressure acc to Raoult's law = 0.9704 * 0.1252 atm = 0.1216 atm
i.e. change = (0.1216 - 0.1252) atm = -0.0036 atm
c. vapor pressure of pure wate r is 25.756 mmHg
vapor pressure reduction of 2.000 mmHg needed
Hence, the changed vap pressure = 23.756 mm Hg
Let's assume x g glucose is needed, so moles of glucose = x/180.2 mol
Water = 250 g= 250/18 mol = 13.89 mol
Hence, mole fraction of water in the solution = 13.89/(13.89 + x/180.2)
So, acc to Raoult's law, [13.89/(13.89 + x/180.2)] * 25.756 = 23.756
i.e. [13.89/(13.89 + x/180.2)] = 0.922
i.e. 13.89 = 12.806 + 5.12 * 10^-3 x
i.e. 1.084 = 5.12 * 10^-3 x Hence, x = 211.6 g ~ 211 g