Question

Calculate the elevation in boiling point for $12g$ of $A{B}_2$ dissolved in $1kg$ of water as solvent if $A{B}_2$ is fully ionised?
Given: $A{B}_2$ dissociates as:
$A{B}_2\to A^{2+}+2B^{-}$
($K_b=0.52Kkgmo{l}^{-1}$
Molar mass of $A{B}_2=120g/mol$)

• A. $0.156K$
• B. $3.141K$
• C. $4.667K$
• D. $0.003K$

Answer 1

The correct option is A $0.156K$

Given,
$A{B}_2$ is an electrolytic solute and it is fully ionised
$A{B}_2\to A^{2+}+2B^{-}$
Thus, the number of ions in the solution is 3
$\therefore$ i = 3, assuming complete ionization of $A{B}_2$
We know,
Corrected boiling point elevation of solution is,
$△T_b=i\times K_b\times m$
where,
$△T_b$ is the elevation in boiling point.
$K_b$ is molal elevation constant.
m is molality of solution.
i is van't Hoff factor
$\therefore$
$△T_b=i\times K_b\times \frac{w_B\times 1000}{m_B\times w_A}$
where,
$w_B$ is mass of solute
$m_B$ is molar mass of solute
$w_A$ is mass of solvent
$△T_b=3\times 0.52\times \frac{12\times 1000}{120\times 1000}=0.156$