Question

How many grams of potassium chloride ($KCl$) must be added to make 500 mL of 1.00M $KCl$ solution?

• A. 1.00 g
• B. 18.6 g
• C. 37.3 g
• D. 74.5 g
• E. 149 g

Answer 1

Answer: (C)

37.3 g

1000 ml 1 M $KCL$ contains 74.5 grams of $KCl$

1 ml 1 M $KCL$ contains $\frac{74.5}{1000}$ grams of $KCl$

Thus, 500 ml 1 M $KCL$ contains $\frac{74.5}{1000}$ $\times$ $500$ grams of $KCl$

Its value comes out to be almost $37.3g$

Thus option $C$ is the correct answer.