A) 1.68
B) 3.36
C) 8.40
D) 0.840
Correct Answer: D
Solution :
Boiling point of solution =boiling point \[+\Delta {{T}_{b}}=100+\Delta {{T}_{b}}\] |
Freezing point of solution = freezing |
Point \[-\,\Delta {{T}_{f}}\] \[=0-\Delta {{T}_{f}}\] |
Difference in temperature (given) \[=100+\Delta {{T}_{b}}-\,(-\,\Delta {{T}_{f}})\] |
\[104=100+\Delta {{T}_{b}}+\Delta {{T}_{f}}\] |
=100 + molality \[\times \,\,{{K}_{b}}\]+molality \[\times \,\,{{K}_{f}}\] |
=100+molality (0.52+1.86) |
\[\therefore \] Molality \[=\frac{104-100}{2.38}=\frac{4}{2.38}=1.68\,\,m\]and molality \[=\frac{Moles\times 1000}{{{W}_{gm\,\,(solvant)}}}\,\,;\] |
\[1.68=\frac{Moles\times 1000}{500}\] |
\[\therefore \]Moles of solute \[=\frac{1.68\times 500}{1000}\]= 0.84 moles |
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