How many grams of NaOH are needed to prepare 250 mL of 0.205 M NaOH?

Question

93896 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-28T07:44:06

A bit over $2*g$ ...............

$"Concentration"$ $=$ $"Moles of solute (moles)"/"Volume of solution (L)"$

Thus $"Concentration"xx"volume"="moles of solute"$

$0.205*mol*cancel(L^-1)xx0.250*cancelL="Moles of solute"$

$=0.0513*mol$ , an answer in moles as required.

And thus $"mass of NaOH"$ $=$ $"moles "xx" molar mass"$

$=0.0513*cancel(mol)xx40.0*g*cancel(mol^-1)$

$=2.05*g$