Question

A chemist wishes to decrease the vapor pressure of a large volume of water. Which amount of solute will decrease it the most? (Note: Assume each sample of solute completely dissolves in the water.)
A)0.10 mol Al2(SO4)3

B)0.10 mol sucrose

C)0.40 mol glucose

D)0.20 mol NaCl

Answer 1

Answer: A) 0.10 mol $Al_2(SO_4)_3$

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,

$\frac{p^o-p_s}{p^o}=i\times x_2$

where,

$\frac{p^o-p_s}{p^o}$= relative lowering in vapor pressure

i = Van'T Hoff factor  

$x_2$ = mole fraction of solute

1. For 0.10 mol $Al_2(SO_{4})_3$

$Al_2(SO_4)_3\rightarrow 2Al^{3+}+3SO_4^{2-}$  

, i= 5 as it is a electrolyte and dissociate to give 5 ions. and concentration of ions will be $2\times 0.1+3\times 0.1=0.5$

2.  For 0.10 mol sucrose

, i= 1 as it is a non electrolyte and does not dissociate, concentration of ions will be $1\times 0.1=0.1$

3. For 0.40 mol glucose

, i= 1 as it is a non electrolyte and does not dissociate,  concentration of ions will be $1\times 0.4=0.4$

4. For 0.2 $NaCl$

$NaCl\rightarrow Na^{+}+Cl^{-}$  

, i= 2 as it is a electrolyte and dissociate to give 2 ions, concentration of ions will be $1\times 0.2+1\times 0.2=0.4$

Thus as concentration of solute is highest for $Al_2(SO_4)_3$ , the vapor pressure will be lowest.

Answer 2

Answer:c

Explanation: