What molar quantity of HCl is there in a 24.38*mL volume of an aqueous solution that is 0.100*mol*L^-1 with respect to HCl?

Apr 25, 2017

Approx. $25 \times {10}^{-} 4 \cdot m o l$.

Explanation:

We use the relationship:

$\text{Concentration}$ $=$ $\text{Moles of solute"/"Volume of solution}$,

And typically, because we often use units of $\text{litres}$ to express volume, $\text{concentration}$ has units ${\text{moles"*"L}}^{-} 1$.

And thus to assess the number of moles in a $24.38 \cdot m L$ volume of $0.100 \cdot m o l \cdot {L}^{-} 1$ $H C l$, we simply assess the product:

$\text{Moles of solute}$ $=$ $\text{Concentration "xx" volume}$

$= 24.38 \times {10}^{-} 3 \cdot \cancel{L} \times 0.100 \cdot m o l \cdot \cancel{{L}^{-} 1} = 24.38 \times {10}^{-} 4 \cdot m o l$.

And we need an equimolar quantity of $N a O H$ to neutralize this molar quantity.