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. $6.023\times {10}^{19}$
• E. $6.023\times {10}^{18}$

Answer 1

The correct option is E $6.023\times {10}^{18}$

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

$\because 100gCaC{O}_3$ gives, molecules of $C{O}_2$ $=6.022\times {10}^{23}$

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

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

$=6.022\times {10}^{18}$.