A) \[{{0.85}^{o}}C\]
B) \[-{{3.53}^{o}}C\]
C) \[{{0}^{o}}C\]
D) \[-{{0.35}^{o}}C\]
Correct Answer: B
Solution :
\[\Delta {{T}_{f}}=i\times {{k}_{f}}\times m\] \[_{Ions\,\,at\,\,equilibrium}^{{}}\] \[\underset{1-\alpha }{\mathop{HBr}}\,\xrightarrow{{}}\underset{\alpha }{\mathop{{{H}^{+}}}}\,+\underset{\alpha }{\mathop{B{{r}^{-}}}}\,\] \[\therefore \] Total ions\[=1-\alpha +\alpha +\alpha \] \[=1+\alpha \] \[\therefore \] \[i=1+\alpha \] Given, \[{{k}_{f}}=1.86\,\,K\,\,mo{{l}^{-1}}\] mass of \[HBr=8.1\,\,g\] mass of \[{{H}_{2}}O=100\,\,g\] \[(\alpha )=\]degree of ionization\[=90%\] \[m\](molality) \[=\frac{mass\,\,of\,\,solute/mol.\,\,wt.\,\,of\,\,solute}{mass\,\,of\,\,solvent\,\,in\,\,kg}\] \[=\frac{8.1/81}{100/1000}\] \[i=1+\alpha \] \[=1+90/100\] \[=1.9\] \[\Delta {{T}_{f}}=i\times {{k}_{f}}\times m\] \[=1.9\times 1.86\times \frac{8.1/81}{100/1000}\] \[={{3.534}^{o}}C\] \[\Delta {{T}_{f}}=\](depression in freezing point) = freezing point of water - freezing point of solution \[3.534=0-\]freezing point of solution. \[\therefore \]Freezing point of solution\[=-{{3.534}^{o}}C\]
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