A) 1.68
B) 3.36
C) 8.40
D) 0.840
Correct Answer: D
Solution :
Boiling point of solution = boiling point \[\Delta \,{{T}_{b}}=100+\Delta {{T}_{b}}\] |
Freezing point of solution = freezing point \[-\Delta \,{{T}_{f}}=0-\Delta \,{{T}_{f}}\] |
Difference in temperature (given) \[=100+\Delta \,{{T}_{b}}-(-\Delta \,{{T}_{f}})\] |
\[104=100+\Delta \,{{T}_{b}}+\Delta \,{{T}_{f}}=100+\text{molality}\times {{\text{K}}_{\text{b}}}+\text{molality}\times {{\text{K}}_{f}}\]\[=100+\text{molality}\,\text{(0}\text{.52+186)}\] |
\[\therefore \] Molality \[=\frac{104-100}{2.38}=\frac{4}{2.38}=1.68\,m\] |
And molality \[=\frac{\text{moles}\times \text{1000}}{{{W}_{gm}}_{\,(solvent)}};\] \[\text{1}\text{.68}=\frac{\text{moles}\times \text{1000}}{500}\] |
\[\therefore \] Moles of solute \[=\frac{1.68\times 500}{1000}=0.84\]moles |
You need to login to perform this action.
You will be redirected in
3 sec