Question

Two elements A and B form compounds having molecular formula $A{B}_2$ and $A{B}_4$. When dissolved in $20$ g $C_6{H}_6$, $1$ g of $A{B}_2$ lowers the f.pt. by ${2.3}^{o}C$ whereas $1.0$ g of $A{B}_4$ lowers it by ${1.3}^{o}C$. The $K_f$ for $C_6{H}_6$ is $5.1Kmo{l}^{-1}kg$. Calculate atomic masses of A and B.

• A. $a=12.79,b=21.34$
• B. $a=25.59,b=42.64$
• C. $a=51.1,b=85.28$
• D. None of these

Answer 1

The correct option is B $a=25.59,b=42.64$

$\because \Delta T=\frac{1000\times K_f\times w}{m\times W}$

For $A{B}_2:2.3=\frac{1000\times 5.1\times 1}{m_1\times 20}$

$\therefore m_1=110.86$ ....(i)

For $A{B}_4:1.3=\frac{1000\times 5.1\times 1}{m_2\times 20}$

$\therefore m_2=196.15$ .....(ii)
Let atomic mass of A and B are a and b respectively.

$\therefore a+2b=m_1=110.86$

$a+4b=m_2=196.15$

$\therefore a=25.59$ and $b=42.64$