15. An aqueous solution contains 34% HCl by mass. A.Calculate the molality of th
ID: 785865 • Letter: 1
Question
15. An aqueous solution contains 34% HCl by mass.
A.Calculate the molality of the solution.
B.Calculate mole fraction of the solution.
16. Calculate the vapor pressure of a solution containing 27.3 g of glycerin (C3H8O3) in 134mL of water at 30.0 ?C. The vapor pressure of pure water at this temperature is 31.8 torr. Assume that glycerin is not volatile and dissolves molecularly (i.e., it is not ionic) and use a density of 1.00 g/mL for the water.
17. A glucose solution contains 53.5 g of glucose (C6H12O6) in 470g of water.
A. Compute the freezing point and boiling point of the solution. (Assume a density of 1.00 g/mL for water.)
B. Compute the freezing point and boiling point of the solution. (Assume a density of 1.00 g/mL for water.)
18. An aqueous solution containing 18.9 g of an unknodn molecular (nonelectrolyte) compound in 107.5g of water was found to have a freezing point of -2.0?C.
A. Calculate the molar mass of the unknown compound.
Explanation / Answer
15) 34 % means 34 gm HCl in 100 gm soln ,
A) molality = (nass/molar mass) /mass of solvent in kg ) (34/35.5) x ( 1000/66) = 14.5 m
B) moles of HCl = 34/35.5 = 0.958 , water moles = 66/18 = 3.667 ,
mol fraction of HCl = (0.958/0.958+3.667) = 0.207 , ml fc of water =1-0.207 = 0.793
16) mol of glycerin = 27.3/92 = 0.297, water vol = 134 ml = 134 gm (since density = 1gm/ml) ,
moles of water = 134/18 = 7.445 , mol fc of solute = (0.297/0.297+7.445) = 0.0384,
(Po-Ps)/Po = Xb , 31.8-Ps = 31.8 x 0.0384 , Ps = 30.58 torr
17) 53.5 gm glucose = 53.5/180 = 0.297, water moles = 470 gm .
molality = (0.297/0.47) = 0.632
A) dT = Tb(soln)-Tb(solvenT) = Kb x m
Tb(soln)-100 = 0.512 x 0.632 , Tb(soln) = 100.324,
dT= Tf(solvent)-Tf(soln) = Kf x m
0-Tf(soln) = 1.86x0.632 , Tf (son) = -1.176 C
B) same as doen in A ( as density mentioned is same)
18) dT = 0-(-2) = 2 = Kf x m
2 = 1.86 xm , m = 1.075 ,
m = 1.075 = ( mass of solute /molar mass) / (solvent mass in kg)
1.075 = ( 18.9/Mol mass) x ( 1000/107.5)
Mol mass = 163.55 gm/mol
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