How many grams of potassium sulfate must be dissolved in 695 grams of acetic acid in order for the boiling point of the solution to be 135.5 decrees Celsius?
mass of solute (K2SO4) = W = ?
moles K2SO4 = mass in grams /molar mass K2SO4 = W / 174.259
mass of solvent (acetic acid) = 695 g = 695/1000 = 0.695 Kg
boiling point of solution = 135.5 C
delta Tb = boiling point of solution - boiling point of pure solvent (acetic acid) = 135.5 -118.1 = 17.4 C
delta Tb = Kb * molality
molality = delta Tb /Kb (acetic acid) = 17.4 / 3.07 = 5.67 m
molality = moles K2SO4 / weight of acteic acid in Kg
put values of moles K2SO4 and mass of acetic acid from above
5.67 = W /174.259 *0.695
W = 5.67 *174.259 *0.695 = 686.69 g =686.7 g (answer)
grams of potassium sulfate dissolved=686.7 g (answer)
please check solution..any doubts let me know..regards ..Sumit sans
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