Respuesta :
Answer: The molal concentration of KOH (aq) is 0.8
Explanation:
Depression in freezing point:
[tex]T^o_f_t_f=i\times k_f\times m[/tex]
where,
[tex]T_f[/tex] = freezing point of solution
[tex]T^o_f[/tex] = freezing point of solvent
[tex]k_f[/tex] = freezing point constant
m = molality
i = Van't Hoff factor = number of ions produced on dissociation
For [tex]KOH[/tex] i =2 as [tex]KOH\rightarrow K^++OH^-[/tex]
For [tex]Al(NO_3)_3[/tex] i =4 as [tex]Al(NO_3)_3\rightarrow Al^{3+}+3NO_3^-[/tex]
Now put all the given values in the above formula, we get:
For 0.4 [tex]Al(NO_3)_3[/tex]
[tex]\Delta T_f=4\times k_f\times 0.4[/tex]
[tex]\Delta T_f=1.6\times k_f[/tex]
For KOH
[tex]\Delta T_f=2\times k_f\times m[/tex]
[tex]1.6\times k_f=2\times k_f\times m[/tex]
[tex]m=0.8[/tex]
Thus molal concentration of KOH (aq) is 0.8
