Freezing point of an aqueous solution is `(-0.186)^(@)C`. Elevation of boiling point of the same solution is `K_(b)=0.512^(@)C`, `K_(f)=1.86^(@)C`. Fi

Question

Freezing point of an aqueous solution is `(-0.186)^(@)C`. Elevation of boiling point of the same solution is `K_(b)=0.512^(@)C`, `K_(f)=1.86^(@)C`. Fi

Freezing point of an aqueous solution is `(-0.186)^(@)C`. Elevation of boiling point of the same solution is `K_(b)=0.512^(@)C`, `K_(f)=1.86^(@)C`. Find the increses in boiling point.
A. `0.186^(@)C`
B. `0.0512^(@)C`
C. `0.092^(@)C`
D. `0.2372^(@)C`

Monjur Alahi

6 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - b
`(DeltaT_(b))/(DeltaT_(f))=K_(b)/K_(f)orDeltaT_(b)=K_(b)/K_(f)xxDeltaT_(f)`.
`DeltaT_(b)=((0.512^(@)C))/((1.86^(@)C))xx(0.186^(@)C)`
`=0.0512^(@)C`

6 views Answered 2023-02-05 15:29