Saved Bookmarks
| 1. |
A solution of glycerol (C_(3) H_(8) O_(3)) in water was prepared by dissolving some glycerol in 500 g of water. This solution has a boiling point of 100.42^(@) C while pure water boils at 100^(@)C. What mass of glycerol was dissolved to make the solution ? (K_(b)"for water " = 0.512 K "kg mol"^(-1)) |
|
Answer» SOLUTION : `(W_(B))` = Mass of GLYCEROL = ? `(W_(A))` = Mass of solvent = 500 g Solution was prepared by dissolving some glycerol in water. (100.42-100) `Delta T_(b)= 0.42^(@)`C `K_(b) = 0.51 " kg mol"^(-1)` `Delta T_(b) = K_(b) xx `Molality. Molality (m) = `(W_(b) xx 1000)/(W_(A) xx M_(B))` Molecular wt.of Glycerol `(M_(B)) C_(3) H_(8) O_(3)` = ` (12 xx 3) + (1 xx 8) + (16 xx 3)` 36 + 8 + 48 = 92 `Delta T_(b) = (W_(B) xx 1000)/(500 xx 92 ) xx K_(b)` `0.42 = (0.512 xx W_(B) xx 1000)/(500 xx 92 )` `W_(B)= (0.42 xx 500 xx 92 )/(0.512 xx 1000)` `W_(B) = (19320)/(512)` `W_(B) `= 37.88 g Weight of glycerol = 37.88 g |
|