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18g of glucose , C6H(12)O6 (Molar mass=1...

18g of glucose , `C_6H_(12)O_6` (Molar mass=180g `mol^-1`) is dissolved in 1000g (1kg) of water in a sauce pan . At what temparature will water boil at 1.013 bar? `K_b` for water is 0.52K kg `mol^-1` . Water boils at 373.15K at 1.013bar pressure.

Text Solution

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`W_(B) = 18 g , M_(B)` (glucose , `C_(6) H_(12) O_(6)`)
`= 1 80 g mol^(-1)`
`W_(A) = 1 kg, , K_(b)`
`= 0 . 52 L kg mol^(-1)`
& `T_(b) = K_(b) xx (W_(B))/(M_(B)) xx (1)/(W_(B) (kg))`
`= ( 0 . 52 K kg mol^(-1) xx 18 g)/(180 g mol^(-1) xx 1 kg)`
`= 0 . 0 52 K`
`= 0 . 0 52 ""^(@)C`
`:.` Boiling point of solutions ,
`T_(b) = T_(b)^(@) + Delta T_(b)`
`= 100 "" ^(@) C + 0052 "" ^(@) C `
`= 100 .052 ""^(@) C `
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