A) \[2.13\text{ }m\]
B) \[0.213\text{ }m\]
C) \[41.66\text{ }m\]
D) \[0.041\,\,m\]
Correct Answer: A
Solution :
\[{{X}_{{{C}_{2}}{{H}_{5}}OH}}=0.04\] \[\therefore \]\[{{X}_{{{H}_{2}}O}}=I-0.04=0.96\] \[\frac{{{X}_{{{C}_{2}}{{H}_{5}}OH}}}{{{X}_{{{H}_{2}}O}}}=\frac{{{n}_{{{C}_{2}}{{H}_{5}}OH}}}{{{n}_{{{H}_{2}}O}}}=\frac{0.04}{0.96}=\frac{1}{24}\] Molality \[=\frac{{{n}_{{{C}_{2}}{{H}_{5}}OH}}}{W(kg)}=\frac{1mol}{\frac{(24\times 18)}{1000}kg}\,\]
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