A) 750.6 mmHg
B) 755.8 mmHg
C) 745.98 mmHg
D) 739.56 mmHg
Correct Answer: C
Solution :
[c] \[\Delta {{p}_{theo.}}=\]Lowering in vapour pressure when there is no dissocation \[={{p}_{0}}\times \frac{wM}{Wm}\] (given, \[{{p}_{0}}=760mm\], w=14g, W=200gm M=18, m=164) \[=\frac{760\times 14\times 18}{200\times 164}=5.84mm\] Degree of dissociation \[\alpha =\frac{70}{100}=0.7\] \[Ca{{(N{{O}_{3}})}_{2}}\rightleftharpoons C{{a}^{2+}}+2N{{O}^{-}}_{3}\] \[(n=3)\] \[i=1+(n-1)\alpha \] \[=1+(3-1)0.7=2.4\] Also, \[i=\frac{\Delta {{p}_{obs.}}}{\Delta {{p}_{theo.}}}\] \[=\frac{\text{No}\text{. of particles after dissociation}}{\text{No}\text{. of particles when there is no dissociation}}\]So. \[\Delta {{p}_{obs}}=2.4\times \Delta {{p}_{theo.}}=2.4\times 5.84\] \[=14.02mm\] \[{{p}_{0}}-{{p}_{s}}=\Delta {{p}_{obs}}=14.02\] \[{{p}_{s}}={{p}_{0}}-14.02=760-14.02=745.98mmHg\]
You need to login to perform this action.
You will be redirected in
3 sec
