11.544 g of a non-volatile solute is dissolved in 235.0 g of water. The solute d

Question

11.544 g of a non-volatile solute is dissolved in 235.0 g of water. The solute does not react with water nor dissociate in solution. Assume that the resulting solution displays ideal Raoult's law behaviour. At 60°C the vapour pressure of the solution is 147.19 torr. The vapour pressure of pure water at 60°C is 149.40 torr. Calculate the molar mass of the solute (g/mol) 58.9 g/mol See example 17.1 on pp718-9 of Zumdahl "Chemical Principles" 8th ed 1pts You are correct. Previous Tries Your receipt no. is 152-7393 Now suppose, instead, that 11.544 g of a volatile solute is dissolved in 235.0 g of water. This solute also does not react with water nor dissociate in solution The pure solute displays, at 60°C, a vapour pressure of 14.94 torr. Again, assume an ideal solution. If, at 60°C the vapour pressure of this solution is also 147.19 torr. Calculate the molar mass of this volatile solute. cf p 719 of Zumdahl "Chemical Principles" 8th ed 1pts Submit Answer Tries 0/5

Explanation / Answer

pTotal = Xsolute*p0solute + Xsolvent*p0 solvent

   n1 = no of mol of solvent = 235/18

   n2 = no of mol of solute = 11.544/x

ntotal = n1+n2

    Xsolute = nsolute / ntotal = ((11.544/x)/((11.544/x)+(235/18))

   Xsolvent = nsolvent / ntotal = ((235/18)/((11.544/x)+(235/18))

p0solute = vapor pressure of pure solute at this temperature = 14.94 torr

p0solvent = vapor pressure of pure water at this temperature = 149.4 torr

   n1 = no of mol of solvent = 235/18

   n2 = no of mol of solute = 11.544/x

vapor pressure of solution = 147.19 torr

147.19 = ((11.544/x)/((11.544/x)+(235/18))*14.94 + ((235/18)/((11.544/x)+(235/18))*149.4

x = molarmass of solute = 52.91g/mol

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