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The boiling point of pure water is 373. ...

The boiling point of pure water is `373. 15 K.` If 32.5 g of KCN is dissolved in 100 mL of `H _(2) O,` then what is the boiling point of the solution ?
(Assume that molarity = molality)
`( K _(b)` for `H _(2) O = 0. 52 K kg mol ^(-1))`
(Molar mass of `KCN = 65 gm mol ^(-1))`

A

`105. 20 ^(@)C`

B

`100. 52 ^(@)C`

C

`373.67 K `

D

`273.67K`

Text Solution

Verified by Experts

The correct Answer is:
A
`Delta T_(b) = i xx m xx K _(b)`
`2 xx ( 32. 5 xx 1000)/( 65 xx 100) xx 0. 52 = Delta T _(b) = 5. 2`
`therefore T _(b) = 373. 15 + 5. 2 = 378 . 35 K `
or `T _(b) = ( 378 . 35 - 273 . 15) = 105 . 20 ^(@) C`
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