Question

$A +2 B +3 C \rightleftharpoons AB _{2} C _{3}$
Reaction of $6.0\, g$ of $A , 6.0 \times 10^{23}$ atoms of $B$, and $0.036\, mol$ of $C$ yields $4.8\, g$ of compound $AB _{2} C _{3}$. If the atomic mass of $A$ and $C$ are 60 and 80 amu respectively, the atomic mass of $B$ is (Avogadro no. $=6 \times 10^{23}$ ):

• A. 70 amu
• B. 60 amu
• C. 50 amu
• D. 40 amu

Answer 1

Answer: (C)

Atomic mass of $A =60$ and atomic mass of $C =80$



Here $C$ is the limiting reagent, hence, $0.036$ mol of $C$ will completely consume $\frac{0.036}{3}=0.012 mol$ of $AB _{2} C _{3}$ and will form $4.8\, gm\, mol$ of

$\left( AB _{2} C _{3}\right)=\frac{\text { mass }}{\text { mol. mass }}$

$0.012=\frac{4.8}{60+M_{ B } \times 2+80 \times 3}$

$\Rightarrow 300+2 M _{ B }=\frac{4.8}{(0.012)}=400$

$\Rightarrow 2 M_{ B }=100$

$\Rightarrow M_{ B }=50\, amu$