A) 746.3 mm Hg
B) 1492.6 mm Hg
C) 373.2 mm Hg
D) 74.63 mm Hg
Correct Answer: A
Solution :
Total mole of \[\operatorname{Ca} {{\left( N{{O}_{3}} \right)}_{2}}\] \[N=(1+2\,\alpha )\frac{7}{164}\] \[=\,\,\,\left( 1+2\times 0.7 \right)\times \,\,\frac{7}{164}\,\,=\,\,0.1\] Molecular weight of \[Ca{{\left( N{{O}_{3}} \right)}_{2}}=164\] \[\operatorname{Mole} of solvent \left( {{H}_{2}}O \right)\,\,=\,\,\frac{100}{18}\] From Raoult?s law, \[\frac{{{p}^{{}^\circ }}-{{p}_{s}}}{{{p}^{{}^\circ }}}\,\,=\,\,\frac{n}{n+N}\] \[\frac{760-{{p}_{s}}}{760}\,\,=\,\,\frac{0.1}{0.1+\frac{100}{18}}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,{{p}_{s}} =746.3 mm Hg\]You need to login to perform this action.
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