Question

Follow the flow of solving like the example given above.You may use the Kb and Kf of water in the example.

1.calculate the freezing and boiling point of a solution prepared by dissolving 15.5 g of Al(NO3)3 in 200.0 g of water.(Molar mass of Al(NO3)3 is 212.996 g/mol).

2. A solution is prepared by dissolving 120 grams of NaCl in 450 grams of water.Find the freezing and boiling point of this solution.(Molar mass of NaCl is 58.44 g/ mol).

Answer 1

$Al(NO 3 ​ ) 3 ​ ⇒Al 3^+ +3NO 3^- ​$

Now here the mole of Aluminium Nitrate dissociates in 4 moles ,hence the value of (i=4)

$Formula used \Rightarrow delta T_f=i\times K_F \times m$

$where K_f=$ 1.86 degre celcius kg mol^-1

m=molality

Now :

Moles of aluminium nitrate

$\frac{15.5}{212.996} ​ =0.0728 moles$

$m=\dfrac{0.0728}{200}\times1000=0.364$

Now: For Freezing point:

$delta T_f=i\times K_f \times m$

$=4\times 1.86\times0.364=2.70$°C

Now we know that

$delta T_f={T^0}_f-T_f$

$Putting {T}^0_f=0$

We get the value of

$T_f=-2.70$ degree celcius and this is our answer