1. What is the temperature at which the solution of KI will freeze if the solution contains 46.83 grams of KI dissolved in 687.4 grams of water. Kf of water is 1.86 C/m.
Write answer to four significant figures.
deltaT_f = K_f*m*i
KI is ionic so i = 2?
m = mol solute/kg solvent = 0.2821118 mol KI / 0.6874 kg H2O = 0.41040418 m
To get mol KI:
46.83 g KI * (1 mol KI/165.998 g KI) = 0.2821118 mol KI
Back to first formula:
x - 0 C = (1.86 C/m)*2*(0.41040418 m)
x = 1.52670355 C = 1.527 C???
2. Calculate the vapor pressure of a solution containing 43.3 grams of CH3OH (P*=140 mm Hg) and 31.6 grams of CH3CH2OH (P*=83.5 mm Hg).
Use P_total = X_a(P_a) + X_b*(P_b) ??
First, CH3OH:
43.3 g CH3OH * (1 mol/32.04 g) = 1.3514357 mol CH3OH
Now, CH3CH2OH:
31.69 g CH3CH2OH * (1 mol/46.07 g ) = 0.687866 mol CH3CH2OH
Find mol fractions of each:
X_CH4O = (1.3514357 mol CH3OH)/(0.687866 mol CH3CH2OH + 1.3514357 mol CH3OH) = 0.66269523
Thanks.