InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The boiling point of benzene is 353.23 K. When 1.80 gm of a nonvolatile silute was dissolved in 90 gm of benzene, the boiling point is raised to 354.11 K. The molar mass of the solute is [`K_(b)` for benzene `= 2.53 K mol^(-1)`]A. `5.8 g mol^(-1)`B. `0.58 g mol^(-1)`C. `58 g mol^(-1)`D. `0.88 g mol^(-1)` |
|
Answer» Correct Answer - C The elevation `(Delta T_(b))` in the boiling point `= 354.11 K - 353.23 K = 0.88 K` Substituting these values in expression `M_("Solute")=(K_(b)xx1000xxw)/(Delta T_(b)xx W)` Where, w = weight of solute, W = weight of solvent `M_("solute")=(2.53xx1.8xx1000)/(0.88xx90)=58 gm mol^(-1)` Hence, molar mass of the solute `= 58 gm mol^(-1)` |
|