What is the molarity of a solution containing 12.0 g of $NaOH$ in 250.0 mL of solution?

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Answer 1 · 2016-05-18T07:49:41

$"Molarity "="Moles of solute"/"Volume of solution"$ $=$ $(12.0*g)/(40.00*g*mol^-1)xx(1)/(250.0xx10^-3*L)$ $=$ $??mol*L^-1$

$"Moles of sodium hydroxide"$ $=$ $(12.0*g)/(40.00*g*mol^-1)$ .

$"Volume of solution"$ $=$ $250.0xx10^-3*L$

We assume (reasonably) that the sodium hydroxide dissolves in the given volume without affecting the volume.

What is the molarity of a solution containing 12.0 g of NaOHNaOH in 250.0 mL of solution?