Question

A substance $A_{2}B$ has composition (by mass) of $40\%A$ and $60\%B$ . Then composition (by mass) of compound $AB_{2}$ is:

• A. $\frac{100}{7}\%A;\frac{600}{7}\%B$
• B. $20\%A;80\%B$
• C. $25\%A;75\%B$
• D. $\frac{100}{11}\%A;\frac{1000}{11}\%B$

Answer 1

Answer: (A)

Here, $\frac{40 / M_{A}}{60 / M_{B}}=\frac{2}{1}\Rightarrow \frac{40}{60}\times \frac{M_{B}}{M_{A}}=\frac{2}{1}\Rightarrow M_{B}=3M_{A}$
$\therefore InAB_{2}:\%$ of $A=\left(\frac{M_{A}}{M_{A} + 2 M_{B}}\right)\times 100\%=\left(\frac{M_{A}}{M_{A} + 6 M_{A}}\right)\times 100\%$
$=\frac{1}{7}\times 100\%=\frac{100}{7}\%$
and $\%$ of $B=\left(100 - \frac{100}{7}\right)\%=\frac{600}{7}\%$