What molar quantity of $H C l$ $HCl$ is there in a $24.38 \cdot m L$ $24.38mL$ volume of an aqueous solution that is $0.100 \cdot m o l \cdot L^{- 1}$ $0.100mol*L^-1$ with respect to $H C l$ $HCl$ ?

Question

What molar quantity of $H C l$ $HCl$ is there in a $24.38 \cdot m L$ $24.38mL$ volume of an aqueous solution that is $0.100 \cdot m o l \cdot L^{- 1}$ $0.100mol*L^-1$ with respect to $H C l$ $HCl$ ?

1 Answer

Impact of this question

2426 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-25T15:48:26

Approx. $25 \times 10^{- 4} \cdot m o l$ $25xx10^-4*mol$ .

Explanation:

We use the relationship:

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

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

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

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

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

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

Science

Math

Humanities

... and beyond

What molar quantity of $H C l$ $HCl$ is there in a $24.38 \cdot m L$ $24.38mL$ volume of an aqueous solution that is $0.100 \cdot m o l \cdot L^{- 1}$ $0.100mol*L^-1$ with respect to $H C l$ $HCl$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

What molar quantity of $H C l$ $HCl$ is there in a $24.38 \cdot m L$ $24.38mL$ volume of an aqueous solution that is $0.100 \cdot m o l \cdot L^{- 1}$ $0.100mol*L^-1$ with respect to $H C l$ $HCl$ ?