Question

To prepare a $0.50$ M KCl solution starting with $100$ mL of $0.40$ M KCl we need to add:

• A. $0.75g$ of KCl
• B. $20mL$ of water
• C. $0.1mol$ of KCl
• D. $0.2mol$ of KCl

Answer 1

Answer: (A)

$0.75g$ of KCl

$Molarity=\frac{n}{V}=\frac{n}{0.1}=0.5$

$\Rightarrow n=0.05$

weight of $KCl=0.05\times 74.5=3.725$

Weight of $KCl$ to be added $=0.01\times 74.5=0.745gm$ which is approximately to $0.75g$.