Question

1.     Calculate the freezing and boiling points 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). 

Answer 1

The reaction takes place as:

$Al(NO_3)_3\Rightarrow Al^{3+}+3{NO_3}^-$

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=$\dfrac{weight}{molar weight}$

= $\dfrac{15.5}{212.996}=0.0728$ moles

Hence :

Molality(m)=$\dfrac{moles}{weight of solvent(gm)}\times1000$

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

Now: For Freezing point:

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

$=4\times 1.86 \times0.364=2.70$ degre celcius

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.

Now for the boiling point

delta$T_b=i\times K_b \times m$

$=$ $4\times 0.512 \times0.364=0.74$ degre celcius

Now:

delta $T_b=T_b-{T}^0_b$

$T_b=0.714+100=100.714$ degree celcius is our answer