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Grace [21]
3 years ago
7

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?

Chemistry
2 answers:
Semmy [17]3 years ago
8 0

<em>Answer: </em>

1) -65.9

2) 100.31

<em> Explanation: edg 2020</em>

<em></em>

MissTica3 years ago
5 0
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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2 years ago
How many grams of H2SO4 are needed to prepare 500. mL of a .250M solution?
zavuch27 [327]

Answer:

We need 12.26 grams H2SO4

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