Question

What molal concentration of KOH(aq) has the same freezing point as 0.4 m Al(NO3)3(aq)? Enter only a numerical answer. Do not include units.

Answer 1

Answer: The molal concentration of KOH (aq) is 0.8

Step-by-step explanation:

Depression in freezing point:

`T^o_f_t_f=i* k_f* m`

where,

`T_f` = freezing point of solution

`T^o_f` = freezing point of solvent

`k_f` = freezing point constant

m = molality

i = Van't Hoff factor = number of ions produced on dissociation

For
`KOH` i =2 as
`KOH\rightarrow K^++OH^-`

For
`Al(NO_3)_3` i =4 as
`Al(NO_3)_3\rightarrow Al^(3+)+3NO_3^-`

Now put all the given values in the above formula, we get:

For 0.4
`Al(NO_3)_3`

`\Delta T_f=4* k_f* 0.4`

`\Delta T_f=1.6* k_f`

For KOH

`\Delta T_f=2* k_f* m`

`1.6* k_f=2* k_f* m`

`m=0.8`

Thus molal concentration of KOH (aq) is 0.8