Solutions Question 83
Question: How many grams of sucrose (M. wt. = 342) should be dissolved in 100 g water in order to produce a solution with a $ 105.0{}^\circ C $ difference between the boiling point and the freezing temperatures- $ (K_{f}=1.86C\text{/}m,K_{b}=0.51{}^\circ C\text{/}m) $
Options:
A) 34.2 g
B) 72 g
C) 342 g
D) 460 g
Show Answer
Answer:
Correct Answer: B
Solution:
$ T_{b}-T_{f}=105=100+5 $
$ T_{b}-T_b^{0}-T_{f}=5 $
$ [\because T_b^{o}=100{}^\circ C] $
$ \Delta T_{b}-T_{f}+T_f^{o}-T_f^{o}=5 $
$ \Rightarrow \Delta T_{b}+\Delta T_{f}=5 $
$ [\because T_f^{o}=0{}^\circ C] $
$ \Delta T_{f}+\Delta T_{b}=m(k_{f}+k_{b}) $
$ \Delta T_{f}+\Delta T_{b}=5=\frac{x}{\frac{342}{\frac{100}{1000}}}(1.86+0.51) $
$ 5=\frac{10x}{342}\times 2.37\Rightarrow x=72,g $
BETA
