A) \[10g\]
B) \[11.4g\]
C) \[9.8g\]
D) \[12.8g\]
Correct Answer: A
Solution :
: According to Raoulfs law: \[\frac{{{p}_{\upsilon }}-{{p}_{s}}}{{{p}_{\upsilon }}}=\frac{{{\omega }_{A}}/{{m}_{A}}}{\frac{{{\omega }_{A}}}{{{m}_{A}}}+\frac{{{\omega }_{B}}}{{{m}_{B}}}}\] where, \[{{\omega }_{A}}\], \[{{m}_{A}}\to \]weight dissolved and molecular mass of solute respectively. \[{{\omega }_{B}},{{m}_{B}}\to \] weight dissolved and molecular mass of solvent respectively. \[{{p}_{\upsilon }}\to \] vapour pressure of pure solvent \[{{p}_{s}}\to \] vapour pressure of solution. \[\frac{{{p}_{0}}-{{p}_{s}}}{{{p}_{\upsilon }}}=\frac{20}{100}=0.2\] \[0.2=\frac{\frac{{{\omega }_{A}}}{40}}{\frac{{{\omega }_{A}}}{40}+\frac{114}{114}}\Rightarrow \frac{{{\omega }_{A}}/40}{\frac{{{\omega }_{A}}+40}{40}}\] \[0.2=\frac{{{\omega }_{A}}}{{{\omega }_{A}}+40}\] \[0.2={{\omega }_{A}}+8={{\omega }_{A}}\] \[8={{\omega }_{A}}-0.2{{\omega }_{A}}=0.8{{\omega }_{A}}\] \[{{\omega }_{A}}=10g\].
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