Question

Calculate the mass of KI in grams required to prepare 5.00 X10^2 mL of a 2.80 M solution

Answer 1

Volume in liters:

5.00x10² mL / 1000 => 0.5 L

Molar mass KI => 166.0028 g/mol

Mass KI = volume x molar mass x molarity

Mass KI = 0.5 x 166.0028 x 2.80

= 232.40392 g of KI

hope this helps!

Answer 2

Step-by-step explanation:

Molarity is the number of moles per liter of solution. Whereas number of moles is defines as mass divided by molar mass.

Therefore, Molarity =
`(mass)/(molar mass * volume of solution)`

Since, it is given that molarity is 2.80 M, volume is 500 ml or 0.5 L, and molar mass of KI is 166 g/mol.

Hence, mass =
`Molarity * molar mass * volume of solution}`

=
`2.80 mol/L * 166 g/mol * 0.5 L`

= 232.4 g

Thus, mass of KI is 232.4 g which is required to prepare 5.00 X10^2 mL of a 2.80 M solution.