A) 62%
B) 98%
C) 55%
D) 75%
Correct Answer: D
Solution :
| Molar mass of solute |
| \[\left( {{M}_{B}} \right)=\frac{1000\times {{K}_{f}}\times {{W}_{B}}}{{{W}_{A}}\times \Delta {{T}_{f}}}\]\[\Rightarrow \,\,\,\,{{M}_{B}}=\frac{1000\times 14\times 75.2}{1000\times 7}\] |
| \[{{M}_{B}}=150.4g\,per\,mol\] |
| Actual molar mass of phenol \[=94g/mol\] |
| Now, van't Hoff factor, |
| \[i=\frac{calculated\,molar\,mass}{observed\,molar\,mass}\] |
| \[\therefore \,\,\,\,i=\frac{94}{150.4}0.625\] |
| Dimerisation of phenol can be shown as: |
| \[\begin{matrix} {} & 2{{C}_{6}}{{H}_{5}}OH & & {{({{C}_{6}}{{H}_{5}}OH)}_{2}} \\ Initial & 1 & {} & 0 \\ At\,equilibrium & 1-\alpha & {} & \frac{\alpha }{2} \\ \end{matrix}\] |
| Total number of moles at equilibrium, |
| \[i=1-\alpha +\frac{\alpha }{2}\] |
| \[i=1-\frac{\alpha }{2}\] |
| But, \[i=0.625\], thus, \[0.625=1-\frac{\alpha }{2}\] |
| \[\frac{\alpha }{2}=1-0.625\] |
| \[\alpha =0.75\] |
| Thus, the percentage of phenol that dimerises is 75%. |
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