A) 2.09 M
B) 2.31 M
C) 20.9 M
D) 23.1M
Correct Answer: A
Solution :
\[{{X}_{{{C}_{2}}{{H}_{5}}OH}}\,\,=\,\,0.040\] \[{{X}_{{{H}_{2}}O}}\,\,=\,\,0.96\] \[\frac{{{X}_{{{C}_{2}}{{H}_{5}}OH}}}{{{X}_{{{H}_{2}}O}}\,}\,\,=\,\,\frac{0.04}{0.96}\,\,=\,\,\frac{1}{24}\,\,\,\therefore \,\frac{{{n}_{{{C}_{2}}{{H}_{5}}OH}}}{{{n}_{{{H}_{2}}O}}}\,\,=\,\,\frac{1\,\,mol}{24\,\,mol}\] \[\therefore \] Total wt. of solution \[=\text{ }46+24\times 18=478g\] \[d\,\,=\,\,1~\,\,\,\,\,\,\,\,\,\therefore \,\,\,\text{ }V\,\,=\,\,478\text{ }ml\] \[Molarity\,\,=\,\,\frac{{{n}_{{{C}_{2}}{{H}_{5}}OH}}}{V}\,\,=\,\,\frac{1\times 1000}{478}\,\,=\,\,2.09\,\,M\]
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