A) 271 K
B) 273.15 K
C) 269.07 K
D) 277.23 K
Correct Answer: C
Solution :
\[\Delta {{T}_{f}}={{K}_{f}}\times \frac{w}{m}\times \frac{1000}{W}\] \[\frac{\Delta {{T}_{f2}}}{\Delta {{T}_{f1}}}=\frac{{{m}_{1}}}{{{m}_{2}}}\] Here, \[{{m}_{1}}\] (cane sugar \[{{C}_{12}}{{H}_{22}}{{O}_{11}}\]) \[=342\] \[{{m}_{2}}\] (glucose \[{{C}_{6}}{{H}_{12}}{{O}_{6}}\]) \[=180\]) \[\Delta {{T}_{f1}}=273.15-271\] \[=2.15K\] \[\frac{\Delta {{T}_{f2}}}{2.15}=\frac{342}{180}\] \[\Delta {{T}_{f2}}=4.085K\] So, freezing point of glucose in water \[=273.15-4.085=269.05\text{ }K\]
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