Question
What is the molarity of a solution containing $$10\,g$$ of $$NaOH$$ in $$500\,mL$$ of solution ?
A.
$$0.25\,\,mol\,\,{L^{ - 1}}$$
B.
$$0.75\,\,mol\,\,{L^{ - 1}}$$
C.
$$0.5\,\,mol\,\,{L^{ - 1}}$$
D.
$$1.25\,\,mol\,\,{L^{ - 1}}$$
Answer :
$$0.5\,\,mol\,\,{L^{ - 1}}$$
Solution :
$$\eqalign{
& {\text{No}}{\text{. of moles of}}\,\,NaOH = \frac{{10}}{{40}} = 0.25\,mol \cr
& M = \frac{{0.25 \times 1000}}{{500}} = 0.5\,\,mol\,\,{L^{ - 1}} \cr} $$