Question #186606

1. Calculate the freezing and boiling points of a solution prepared by dissolving

15.5 g of Al (NO_{3})_{3 }in 200.0 g of water. (Molar mass of Al (NO_{3})_{3 }is 212.996 g/mol).

Expert's answer

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

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