Question

The amount of oxalic acid, required to prepare $300mL2.5M$ solution, is:

• A. 67.5 g
• B. 9.45 g
• C. 6.75 g
• D. 94.5 g

Answer 1

Answer: (A)

Molarity $=\frac{\text{ moles of solute }}{\text{ volume of solution in L }}=\frac{\text{ mass/molecular mass }}{\text{ V in L }}$
Given molarity $=2.5M$
Volume $=300mL=\frac{300}{1000}L$
Molecular mass of oxalic acid $(COOH{)}_2$
$=(12+16\times 2+1{)}_2=902.5=\frac{\text{ mass }/90}{300/1000}$
or $2.5=\frac{\text{ mass }}{90}\times \frac{1000}{300}$
mass $=\frac{90\times 300\times 2.5}{1000}=67.5g$