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Boiling point of water at 750 mm Hg is 9...

Boiling point of water at 750 mm Hg is `99.63^(@)C`. How sucrose is to be added to 500 g of water such that it boils at `100^(@)C`. Molal elevation constant for water is 0.52 K kg `mol^(-1)`.

Text Solution

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Here, elevation of boiling point
`Delta T_(b)=(100+273)-(99.63+273)=0.37 K`
Mass of water, `w_(1)=500 g`
Molar mass of sucrose `(C_(12)H_(22)O_(11))`,
`M_(2)=11xx12+22xx1+11xx16=34 g mol^(-1)`
Molal elevation constant,
`K_(b)=0.52 " K kg mol"^(-1)`
We know that :
`Delta T_(f)=(K_(b)xx1000xx w_(2))/(M_(2)xx w_(1))`
`w_(2)=(Delta T_(b)xx M_(2)xx w_(1))/(K_(b)xx1000)=(0.37xx342xx500)/(0.52xx1000)`
= 121.7 (approximately)
Hence, 121.67 g of sucrose is to be added.
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