Saved Bookmarks
| 1. |
A solution containing 0.5216g of naphthalene (mol.wt.=128.16) in 50mL of "CC"l_(4) shows boiling point elevation of 0.402^(@) while a solution of 0.6216g of an unknown solute in the same weight of solvent gave a boiling point elevation of 0.647^(@) . Find the molecular mass of the unknown solute. |
|
Answer» Solution :Suppose the weight of 50mL of solvent `"CC"l_(4)` is W grams. For the FIRST solution , molality `=(0.5216//128.16)/(W)xx1000` `=(521.6)/(128.16W)` `K_(b)=(DeltaT_(b))/(m)` `=(0.402)/(521.6//128.16W)` As the second solution is prepared in the same weight of solvent, so for the second solution molality `=(0.6216//M)/(W)xx1000` `=(621.6)/(WM)` `:.K_(b).=(0.647)/(621.6//WM)` (M is the mol.wt. of the UNKNOWN SOLUTE) Since the solvent in both the solutions is the same `K_(b)=K_(b).` Thus, `(0.402)/(521.6//128.16W)=(0.647)/(621.6//WM)` `M=94.84` |
|