Question

How many molecules of $C{O}_2$ are formed when one milligram of $100\%$ pure $CaC{O}_3$ is treated with excess hydrochloric acid?

• A. $6.023\times 10^{23}$
• B. $6.023\times 10^{21}$
• C. $6.023\times 10^{20}$
• D. $6.023\times 10^{18}$
• E. $6.023\times 10^{19}$

Answer 1

Answer: (D)

$\underset{100g}{CaC{O}_3}+2HCl\to CaC{l}_2+H_{2}O+\underset{\text{1 mol}}{C{O}_2}$

$=6.022\times 10^{23}$ molecules

$\because 100gCaC{O}_3$ gives, molecules of $C{O}_2$

$=6.1022\times 10^{23}$

$\therefore 1\times 10^{-3}gCaC{O}_3$ gives molecules of $C{O}_2$

$=\frac{6.022\times 10^{23}\times 1\times 10^{-3}}{100}$

$=6.022\times 10^{18}$