Question

1. Calculate the molality of a solution of 50mL of ethanol, C2H5OH (density .789g.mL) which is dissolved in .25L of water (density=1.00g/mL).

2. A solution is prepared by diluting 225mL of .1885 M aluminum sulfate with water to a final volume of 1.450L. Calculate the molarities of the aluminum sulfate, aluminum ions, and sulfate ions in the diluted solutions.

Answer 1

1)

M = mol / L

V = 50 ml + 250 ml = 300 ml or 0.3 L

then

mol of C2H5OH = mass/MW

mass = D*V = 0.789*50 = 39.45 g

then

mol of C2H5OH = 39.45 /46 = 0.857608

then

M = mol/L = 0.857608/0.3 = 2.85 M

note that I assume volumes are additive (i.e. volume of alcohol + volume of water = total volume)

if volume are not additive, then

VT = V water = 0.25

then

M = mol/L = 0.857608/0.25 = 3.43

2)

V = 225 ml = 0.225 L

Vf = 1.45

apply dilutoin law

M1V1 = M2V2

M2 = M1V1/V2 = 0.1885*0.225/1.45 = 0.02925 M

this is for Al2(SO4)3

then

1 mol of Al2(SO4)3 --> 2 mol of Al+3 and 3 mol of SO4-2

therefore

for Al+3 = 0.02925 *2 = 0.0585 Al+3

for SO4-2 = 0.02925 *3 =0.08775 M of SO4-2