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

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Answer 1 · 2017-04-25T15:48:26

Approx. $25xx10^-4*mol$ .

We use the relationship:

$"Concentration"$ $=$ $"Moles of solute"/"Volume of solution"$ ,

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

And thus to assess the number of moles in a $24.38*mL$ volume of $0.100*mol*L^-1$ $HCl$ , we simply assess the product:

$"Moles of solute"$ $=$ $"Concentration "xx" volume"$

$=24.38xx10^-3*cancelLxx0.100*mol*cancel(L^-1)=24.38xx10^-4*mol$ .

And we need an equimolar quantity of $NaOH$ to neutralize this molar quantity.

What molar quantity of HClHCl is there in a 24.38*mL24.38*mL volume of an aqueous solution that is 0.100*mol*L^-10.100*mol*L^-1 with respect to HClHCl?