Question

A substance $A_{2}B$ has composition (by mass) of $40\%A$ and $60\%B$ . Then composition (by mass) of compound $A{B}_2$ 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, $\frac{40/M_A}{60/M_B}=\frac{2}{1}\Rightarrow \frac{40}{60}\times \frac{M_B}{M_A}=\frac{2}{1}\Rightarrow M_B=3M_A$
$\therefore InA{B}_2:\%$ of $A=\left(\frac{M_A}{M_A+2M_B}\right)\times 100\%=\left(\frac{M_A}{M_A+6M_A}\right)\times 100\%$
$=\frac{1}{7}\times 100\%=\frac{100}{7}\%$
and $\%$ of $B=\left(100-\frac{100}{7}\right)\%=\frac{600}{7}\%$