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).
____________________
⭐_BTS_Army_fan__❤❤
Attachments:
Answers
Answered by
15
Answer:
HEYY here bts fan ...
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.
hope it helps...❤...
please mark me as a brainliest..❤✌..
Answered by
5
Brother mark she's as Brilliant.
⭐⭐⭐⭐⭐⤴⤴⤴
Similar questions