Answer to Question #254383 in General Chemistry for Rose

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