Question

While combusting in air, $4\, g$ of $H _{2}$ gas was completely converted into water. If $36\, \mu$ mole of $CO _{2}$ from air is dissolved into that water, what is the concentration of $CO _{2}$ ?

• A. $1\, \mu M$
• B. $1\, mM$
• C. $1\, nM$
• D. $1000\, mM$

Answer 1

Answer: (B)

The equation takes place as follows:

$\underset{4g}{2 H _{2}}+ O _{2} \longrightarrow \underset{3 g }{2 H _{2} O }$

$\because 36\, g$ of $CO _{2}$ is dissolved in $36\, g$ of $H _{2} O$.

$\therefore $ Conc. of $CO _{2} =\frac{36 \times 10^{-6} \times 10^{3}}{36}$

$\left(\because \text{conc.} (C)=\frac{n}{V}\right)$

$=10^{-3} M =1\, mM$