A) Double
B) Three times
C) Four times
D) Same
Correct Answer: C
Solution :
\[\therefore \] \[\text{NaCl}=\frac{3.33}{4.44}\times \text{1}00=\text{75}%\] \[pH=p{{K}_{a}}+\log \frac{[salt]}{[acid]}\] \[5.0=4.0+\log \frac{[salt]}{acid}\] \[\frac{[salt]}{acid}=anti\log 1=10\] One mole of \[{{N}_{2}}{{O}_{4}}\]decomposes to gives 2 moles of\[=\frac{12}{44}\times \frac{0.66}{m}\times 100=\frac{18}{m}\]. \[=\frac{2}{18}\times \frac{0.36}{m}\times 100=\frac{4}{m}\] \[={{C}_{3}}{{H}_{8}}\] \[\pi \] \[{{M}_{2}}=\frac{1000{{K}_{t}}\times {{W}_{2}}}{{{W}_{1}}\times \Delta T}=\frac{1000\times 5\times 2.5}{25\times 2.25}=222\] \[\therefore \] \[i=\frac{{{M}_{T}}}{{{M}_{0}}}=\frac{122}{222}=0.5492=0.55\] \[\alpha =\frac{1-i}{1-\frac{1}{x}}-=\frac{1-0.55}{1-\frac{1}{2}}=\frac{0.45}{0.5}=0.9\] = four times the initial pressure.You need to login to perform this action.
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