psoln = XH2O*PoH2O
81.6 = XH2O*92.6
XH2O = 81.6/92.6 = approx 0.88 but you need a more accurate number than that estimate.
Set up mole fraction expression for H2O.
[nH2O/(nH2O + nCaCl2)] = 0.88
nH2O = 1 in the problem;solve for nCaCl2 (as ions) and I get about 0.135. The van't Hoff factor is 3 for CaCl2 so mols CaCl2 in the molecular form is 0.135/3 = about 0.045 and convert that to grams CaCl2. That's approximately 5.0 grams.
% w/w = (mass solute/mass solution)*100 = ?
% = [(5.0/(5.0+ 18)]*100 = about 22% w/w.
Another way to do this is
vapor pressure LOWERING or
delta P = i*XCaCl2*PoH2O
92.6-81.6 = 11 = XCaCl2*92.6 and solve for XCaCl2. XCaCl2 = about 0.12 so XH2O = 1.0-0.12 = about 0.88 and go from there.
Determine the mass(w/w%) percent of CaCl2 given the following information. The vapor pressure of a CaCl2 solution is 81.6 mmHg at 50degrees C. The vapor pressure of pure water at this temp is 92.6mmHg. Assume that you have 1.00mol of water.[Raoult's Law: P(solution)=X(solvent)P^o
(solvent)]
1 answer
1 answer