A) \[12.2\,\,g\,mo{{l}^{-1}}\]
B) 15.4 g mol
C) \[17.3\,\,g\,mo{{l}^{-1}}\]
D) 20.4 g mol
Correct Answer: C
Solution :
\[\operatorname{First} boiling point of water = 100{}^\circ C\] \[\operatorname{Final} boiling point of water = 100.52{}^\circ C\] \[\operatorname{w} =3 g,\,\,W=200 g,\,\,{{K}_{b}}, =0.6 k{{g}^{-}}^{1}\] \[\Delta {{T}_{b}}=100.52-100 =0.052{}^\circ C\] \[m=\frac{{{K}_{b}}\times w\times 1000}{\Delta {{T}_{b}}\times W}\] \[=\,\,\,\frac{0.6\times 3\times 1000}{0.52\times 200}=\frac{1800}{104}=17.3\,\,g\,mo{{l}^{-1}}\]
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