Saved Bookmarks
| 1. |
One hundred gram of a 5 m urea solutionare cooled to- 6^(@)C .Whatamountof urea will seprate out ? (K_(f) = 1.86 K m^(-1)) ? |
|
Answer» Solution :Molalityof solution at ` - 6^(@)C` `Delta T_(f) = k_(f) m` or `"" m = (Delta T_(f))/(K_(f)) = (6)/(1.86) = 3.23 MOL kg ^(-1)` 5 m solution of UREA means that 5 MOLES or `5 xx 60` gof urea are present in 1000 g of solventor `(1000+5xx6)`g of solution ,so that . `(100+300)` g of solutioncontain urea = 300 g . 100g of solutioncontain urea ` = (300)/(1300) xx 100` ` =23.08 g` Mass of water in the solution`= 100- 23.08` ` = 76.92 g` Whenurea separates out , mass of waterdoes notchange.At ` - 6^(@)C`, the molalityis `3.23 mol kg^(-1)` . Letus calculatethe amountof urea to makethe MOLALITY`3.23 mol kg^(-1)` . 1000 g of watercontain urea ` = 60 xx 3.23 g` `76.92` g of watercontain urea ` = (60 xx 3.23 xx 76.92)/(1000)` ` = 14.91 g` Mass ofurea SEPARATED ` = 23.08- 14.91` ` = 8.17g` |
|