A) 0.4
B) 0.3
C) 0.2
D) 0.1
Correct Answer: D
Solution :
Weight of solute (HCl) = 3.65 g Molecular wt. of solute (HCl) \[=1+33.5=36.5\] Weight of solvent \[({{H}_{2}}O)=16.2\,g\] Molecular wt. of solvent \[({{H}_{2}}O)\] \[=2+16=18g\] Mole fraction of \[\text{HCl =}\frac{\text{Moles}\,\text{of}\,\text{HCl}}{\text{Molecular}\,\text{wt}\text{.}\,\text{of}\,\text{HCl}}\] Moles of \[\text{HCl}\,\text{=}\frac{\text{Wt}\text{.of}\,\text{HCl}}{\,\text{Molecular}\,\text{wt}\text{.}\,\text{of}\,\text{HCl}}\] \[=\frac{3.65}{36.5}=0.1\] Moles of \[{{\text{H}}_{\text{2}}}\text{O}\,\text{=}\frac{\text{Wt}\text{.}\,\text{of}\,{{\text{H}}_{\text{2}}}\text{O}}{\text{Molecular}\,\text{wt}\text{.}\,\text{of}\,{{\text{H}}_{\text{2}}}\text{O}}\] \[=\frac{16.2}{18}=0.9\] So, mole fraction of \[\text{HCl}\,\text{=}\,\frac{\text{0}\text{.1}}{\text{0}\text{.1}\,\text{+}\,\text{0}\text{.9}\,}\] \[=\frac{0.1}{1}=0.1\]
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