A) 34.2
B) 171.2
C) 68.4
D) 136.8
Correct Answer: C
Solution :
Since two solutions are isotonic, they must have same concentrations in moles/?. For cane sugar we have, \[\operatorname{Concentration} = 5g/100c{{m}^{3}} \left( given \right)\] \[= 50g/l=\frac{50}{342} moles/l\] \[\left( \therefore \text{ }Molar\text{ }mass\text{ }of\text{ }cane\text{ }sugar=342\text{ }\operatorname{g} mo{{l}^{-1}}\text{ } \right)\] For substance X, let M is the mol. Mass \[\operatorname{Concentration} = lg/100 c{{m}^{3}}\left( given \right)\] \[= 10g/l=\frac{10}{\operatorname{M}} moles/l\] \[\operatorname{Thus} have \frac{10}{M}=\frac{50}{342}\] \[\operatorname{M}=68.4\,\,g\,\,mo{{l}^{-1}}\]
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