Question

How many mL of 0.200M KI would contain 0.0500 moles of KI?

Please explain and show work.

Answer 1

250ml

Step-by-step explanation:

call it V

V*0.2=0.05 (moles)

so V=0.05/0.2 = 0.25l = 250ml

Answer 2

• Molarity=0.2M
• No of moles=0.05mol

We know

`\boxed{\Large{\sf Molarity=(No\:of\:moles\:of\:solute)/(Volume\:of\:solution\:in\;\ell)}}`

`\\ \Large\sf\longmapsto Volume\:of\:KI=(0.05)/(0.2)`

`\\ \Large\sf\longmapsto Volume\:of\:KI=0.25L`

`\\ \Large\sf\longmapsto Volume\:of\:KI=250mL`