Question

3. When 7.45 grams on a nonvolatile organic compound was dissolved in 250 grams of 2- propanol, the vapor pressure was measured to be 48.58 kPa. If the vapor pressure of pure 2- propanol is 50.00 kPa at 333.75 K, what is the molar mass of the nonvolatile organic compound?

Answer 1

Answer

61.27g/mol

Explanation

According to ravoult's law

P_solution = X_solvent P^°_solvent

P_solution = Vapor pressure of solution , 48.58 kPa

X_solvent = mole fraction of solvent at solution , ?

P°solvent = Vapour pressure of solvent in its pure state, 50.00 kPa

X_solvent = P_solution/P°_solvent

X_solvent = 48.58kPa/50.00kPa

X_solvent = 0.9716

mole fraction of 2-propanol = 0.9716

mole fraction of 2-propanol = no of moles of 2- propanol /Total no of moles

No of moles of 2-propanol = 250g/ 60.10g/mol = 4.1597mol

Total no of moles = no of moles of 2-propanol/mole fraction of 2-propanol

Total no of moles = 4.1597mol/0.9716 = 4.2813

moles of nonvolatile compound = Total no of moles - no of moles of 2-propanol

moles of nonvolatile compound = 4.2813 mol - 4.1597mol = 0.1216mol

molar mass = mass/no of moles

Therefore,

Molar mass of nonvolatile organic compound = 7.45g/0.1216mol = 61.27g/mol