plzz solve question no.21
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Initial boiling point:353.23K
final boiling point=254.11K
change in boiling point,ΔT b=354.11-353.23=0.88K
given,weight of solute,w₂=1.8g
weight of solvent,w₁=90g
Kb=2.53
molar mass of solute,M₂=? (to be calculated)
using formula:
ΔTb=(Kb×1000×w₂)/M₂×w₁
form this formula,
M₂=(Kb×1000×w₂)/ΔTb×w₁
=(2.53×1000×1.8)/0.88×90
=4554/79.2
=57.5 g/mol
Since the nearest value is 58g/mol that is the right answer
Hope it helped.!!!
final boiling point=254.11K
change in boiling point,ΔT b=354.11-353.23=0.88K
given,weight of solute,w₂=1.8g
weight of solvent,w₁=90g
Kb=2.53
molar mass of solute,M₂=? (to be calculated)
using formula:
ΔTb=(Kb×1000×w₂)/M₂×w₁
form this formula,
M₂=(Kb×1000×w₂)/ΔTb×w₁
=(2.53×1000×1.8)/0.88×90
=4554/79.2
=57.5 g/mol
Since the nearest value is 58g/mol that is the right answer
Hope it helped.!!!
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