Saved Bookmarks
| 1. |
What would be the molar mass of a compound if I 6.21 g of it dissolved in 24 .0 g of chloroform form a solution that has a bolling point of 68.04""^(@)C. The bolling point of pure chloroform is 61.7^(@)C and the boiling point elevation constant, K_(b) for chloroform is 3.63^(@)C/m. |
|
Answer» Solution :`DeltaT_(b)=K_(b)xx(W_(B))/(M_(B))xx(1000)/(W_(A))` `T_(b)-T_(b)^(0)=(3.63xx6.21xx1000)/(M_(B)xx24)` `61.7-68.04 = (3.63xx6210)/(M_(B)xx24)` `M_(b)=(3.63xx6210)/(24xx6.34)` `=(22542.3)/(152.16)` `=148.148g//"mol"^(-1)` |
|