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A solution of glycerol (C(3) H(8) O(3)) ...

A solution of glycerol `(C_(3) H_(8) O_(3))` in water as prepared by dissolving some glycerol in 500 g of water. This solution has a boiling point of 100 . `42^(@)` C. What mass of glycerol was dissolved to make the solution ? `K_(b) ` for `H_(2) O` = 0 . 512 K kg ` mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
37.7 g
`Delta T_(b) = T_(b) - T_(b)^(@) = (100 .42 - 100)^(@)C`
`= 0.42^(@) C = 0.42 K`
`K_(b) = 0.512 K kg mol^(-)`
`W_(A) = 500 g = 0 . 5 kg`
`n_(B) = ?`
`Delta T_(b) = K_() xx (n_(B))/(W_(A) (kg))`
`n_(B) = (Delta T_(b) xx W_(A) (kg))/( K_(b)`
`= ((0.42 K) xx (0.5 kg))/( (0.512 K kg mol^(-1))`
`= (42 xx 5)/( 512) ` mol 0.41 mol
Molar mass of glycerol,
`C_(3) H_(8) O_(3) = (36 + 8 + 48) g mol^(-1)`
`=92 g mol^(-1)`
`n_(B) = (W_(B))/(M_(B))`
`W_(B) = n_(B) xx M_(B) = 0.41mol 92 g mol ^(-1) = 37 . 7 g `
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Knowledge Check

  • 18 g of glucose is dissolved in 1 kg of water. At what temperature will the solution boil ? ( K_b for water is 0.52 K kg "mol"^(-1) )

    A
    ` 99.052^@C`
    B
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    C
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    D
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