What mass of lithium hydroxide is need to prepare a $250mL$ volume of $"LiOH"$ at $0.33mol*L^-1$ concentration?

Question

10797 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-24T16:32:30

We require an approximate mass of $2.0*g$ of $LiOH$ .

$"Concentration"$ $=$ $"Moles of solute"/"Litres of solution"$ . And thus the typical units of concentration are $mol*L^-1$ .

We want to make a $250*mL$ volume of $0.33*mol*L^-1$ concentration. And thus we need $0.250*Lxx0.33*mol*L^-1$ $=$ $0.0825*mol$ of $LiOH$ solute.

And this molar quantity has a mass of $0.0825*molxx23.95*g*mol^-1$ $=$ $1.98*g$

What mass of lithium hydroxide is need to prepare a 250*mL250*mL volume of "LiOH""LiOH" at 0.33*mol*L^-10.33*mol*L^-1 concentration?