Question

Calculate the boiling point of solution when 2g of Na2SO4 ( M = 142 g/mol )
was dissolved in 50g of water, assuming Na2SO4 undergoes complete ionization.
( Kb for water = 0.52 k kg/mol).
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Answer 1

Answer:

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It's answer is Answer

Given: Wsolute (W2) = 2g

Msolute (M2)= 142 g/mol

Wsolvent (W1) = 50g

Kb for water = 0.52 K kg /mol

Dissociation of Na2SO4:

Na2SO4→ 2Na+ + SO42-

i(no. of particles after dissociation) =3

First, we apply the formula:

ΔTb = i×Kb× m

⇒ ΔTb = 0.439 K

To calculate boiling point of the solution, we apply the formula given below:

Tb = Tb° + ΔTb

⇒ Tb = 100 K + 0.439 K

⇒ Tb = 100.496K

Thus, the boiling point of the solution is 100.496 K.

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Answer 2

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