12.682 g of a non-volatile solute is dissolved in 160.0 g of water. The solute d

Question

12.682 g of a non-volatile solute is dissolved in 160.0 g of water. The solute does not react with water nor dissociate in solution. At 70°C the vapour pressure of the solution is 230.47 torr. The vapour pressure of pure water at 70°C is 233.70 torr. Calculate the molar mass of the solute (g/mol) Submit Answer Tries o/99 Now suppose, instead, that 12.682 g of a volatile solute is dissolved in 160.0 g of waten. This solute also does not react with water nor dissociate in solution. Again, assume an ideal solution. If, at 70°C the vapour pressure of this solution is also 230.47 torr.

Explanation / Answer

raoults law

   P0-P/P0 = i*Xsolute

    P0-P/P0 = n2 / n1+n2

i = vanthoffs factor of solute = 1

P = vapor pressure of water above the solution = 230.47 torr

p0 = vapor pressure of pure water at this temperature = 233.7 torr

   n1 = no of mol of solvent = (160/18) = 8.9 mol

   n2 = no of mol of solute particles = (12.682/x) mol

(233.7-230.47)/233.7 = ((12.682/x)/((12.682/x)+8.9)

x = molarmass of solute = 101.7 g/mol

from daltons law

Ptotal = Psolute + pSolvent

         = Xsolute*P0Solute + Xsolvent*P0solvent

    230.47 = ((12.682/x)/((12.682/x)+8.9)*23.37 + ((8.9)/((12.682/x)+8.9)*233.7

x = molarmass of solute = 91.36 g/mol

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.