Saved Bookmarks
| 1. |
A solution containing 0.5126 g naphthalene (mol. Mass = 128) in 50.0 g of carbon tetrachloride yields a boiling point elevation of 0.402^(@)C while a solution of 0.6216 g of an unknown solute in the same weight of the same solvent gives a boiling point elevation of 0.647^(@)C. Find the molecular mass of the unknown solute. |
|
Answer» SOLUTION :In this case, the first data is USED to find the value of `K_(B)`. This value of `K_(b)` is used in the second data to find the molecular mass. Srep I. To find `K_(b)` for `C Cl_(4)` from data on napthalene solution `w_(2)=0.5126 g, w_(1)=50.00 g, M_(2)=128 g mol^(-1), Delta T_(b)=0.402^(@)C=0.402 K^(**)` `K_(b)=(Delta T_(b).w_(1)M_(2))/(1000 w_(2))=(0.402 K XX 50 g xx 128 g mol^(-1))/(1000 g kg^(-1)xx0.5126 g)=5.02" K kg mol"^(-1)` Steo II. To find the mol. mass of unknown solute. The given data are : `w_(2)=0.6216 g, w_(1)=50.00 g, Delta T_(b)=0.647^(@)C=0.647 K` `K_(b)=5.02 "K kg mol"^(-1)` (calculated above) `M_(2)=(1000 K_(b)w_(2))/(w_(1)Delta T_(b))=(1000 g kg^(-1)xx5.02"K kg mol"^(-1)xx 0.6216 g)/(50g xx0.647 K)=96.46 g mol^(-1)` |
|