Answer: A) 0.10 mol [tex]Al_2(SO_4)_3[/tex]
Explanation:
As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.
The formula for relative lowering of vapor pressure will be,
[tex]\frac{p^o-p_s}{p^o}=i\times x_2[/tex]
where,
[tex]\frac{p^o-p_s}{p^o}[/tex]= relative lowering in vapor pressure
i = Van'T Hoff factor
[tex]x_2[/tex] = mole fraction of solute
1. For 0.10 mol [tex]Al_2(SO_{4})_3[/tex]
[tex]Al_2(SO_4)_3\rightarrow 2Al^{3+}+3SO_4^{2-}[/tex]
, i= 5 as it is a electrolyte and dissociate to give 5 ions. and concentration of ions will be [tex]2\times 0.1+3\times 0.1=0.5[/tex]
2. For 0.10 mol sucrose
, i= 1 as it is a non electrolyte and does not dissociate, concentration of ions will be [tex]1\times 0.1=0.1[/tex]
3. For 0.40 mol glucose
, i= 1 as it is a non electrolyte and does not dissociate, concentration of ions will be [tex]1\times 0.4=0.4[/tex]
4. For 0.2 [tex]NaCl[/tex]
[tex]NaCl\rightarrow Na^{+}+Cl^{-}[/tex]
, i= 2 as it is a electrolyte and dissociate to give 2 ions, concentration of ions will be [tex]1\times 0.2+1\times 0.2=0.4[/tex]
Thus as concentration of solute is highest for [tex]Al_2(SO_4)_3[/tex] , the vapor pressure will be lowest.
