Question

The ppm strength of $C{O}_2\left(g\right)$ volume by volume (mL of $C{O}_2$ per ${10}^6$ mL of air) is:

• A. $224$
• B. $2240$
• C. $100$
• D. $1000$

Answer 1

Answer: (D)

$1000$

Milliequivalents of $KOH$ (total) $=500\times 0.1\times 1=50$

Milliequivalents of $HCl$ for $KOH$ ($50$ mL) $=30\times 0.1=3$

Milliequivalents of excess $KOH$ in $500$ mL $=\frac{3}{50}\times 500=30$

Milliequivalents of $KOH$ reacted $=50-30=20=$ mEq. of $C{O}_2$

Millimoles of $C{O}_2=\frac{20}{2}=10=10\times {10}^{-3}$ mol

($1$ mol of $C{O}_2=22.4$ L at STP)

ppm of $C{O}_2=\frac{\left[\right(10\times {10}^{-3}\times 22.4)\times \text{/}{10}^3]\times {10}^6}{224\times \text{/}{10}^3}$ $=1000$