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:c

Step-by-step explanation:

Answer 2

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

Step-by-step 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,

`(p^o-p_s)/(p^o)=i* x_2`

where,

`(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* 0.1+3* 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* 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* 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* 0.2+1* 0.2=0.4`

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