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How many grams of potassium sulfate must be dissolved in 695 grams of acetic

Chemistry
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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?

Apr 17th, 2015

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)

Apr 17th, 2015

please check solution..any doubts let me know..regards ..Sumit sans

Apr 17th, 2015

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