To observe an elevation of boiling point of `0.05^(@)C`, the amount of a solute (molecular weight = 100) to be added to 100 g of water `(K_(b) = 0.5)`

Question

To observe an elevation of boiling point of `0.05^(@)C`, the amount of a solute (molecular weight = 100) to be added to 100 g of water `(K_(b) = 0.5)`

To observe an elevation of boiling point of `0.05^(@)C`, the amount of a solute (molecular weight = 100) to be added to 100 g of water `(K_(b) = 0.5)` is
A. 2 g
B. 0.05 g
C. 1 g
D. 0.75 g

Monjur Alahi

4 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 - c
Elevation of boiling point,
`DeltaT_(b) = (wxxK_(b)xx1000)/(MxxW_("gram"))`
On substituting values, we get
`0.05 = (wxx0.5xx1000)/(100xxx100)` or `w = (0.05xx100xx100)/(0.5xx1000) = 1g`

4 views Answered 2023-02-05 15:29