Determine the boiling point of a solution made by dissolving 0.60 mol K2SO4 in 1.0 kg water. Water has a boiling point elevation constant of 0.51°C•kg/mol. What is the boiling point of this solution?
2 answers:
<em>Answer:
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1) -65.9
2) 100.31
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Explanation: edg 2020</em>
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To determine the boiling point (bp), you need this formula---> bp solution= bp solvent + Δbp solution bp solvent is known since we are dealing with water; it is 100C all we have to do is to solve for Δbp solution and add it to the 100C. Δbp solution= Kb x i x m, where Kb is the constant, i is the number of particles and m is molality. Kb= 0.51 i= 3, K2SO4 ----> 2K+ + SO4-2 (the solute breaks into three ions) m= ? = moles/kg of solvent molality (m)= 0.60 mol/ 1.0 Kg= 0.60 m Δbp solution= 0.51 x 3 x 0.60= 0.918 C bp solution= 100C + 0.918 C= 100.918 C
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