Question

What is the freezing point ( in c) of a 0.527 m aqueous solution of LiBr enter your rounded answer with 3 decimal places

Answer 1

Answer:-1.96

Step-by-step explanation:

Answer 2

Answer:
`-1.960^o C`.

Step-by-step explanation:

`\Delta T_f=i* K_f* m`

`\Delta T_f` = depression in freezing point

`K_f` = freezing point constant =
`1.86^0C/m`

i= vant hoff factor = number of ions produced on complete dissociation

`LiBr\rightarrow Li^++Br^-`

For
`LiBr` , i= 2

m= molality = 0.527

`\Delta T_f=T_f^0-T_f=(0-T_f)^0C`

`0-T_f=2* 1.86* 0.527`

`T_f=-1.960^0C`

The freezing point of the solution is
`-1.960^o C`.