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

2 answers:

3 0

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!

Evgen [1.6K]2 years ago

3 0

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 = $\frac{mass}{molar mass \times volume of solution}$

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 = $Molarity \times molar mass \times volume of solution}$

= $2.80 mol/L \times 166 g/mol \times 0.5 L$

= 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.

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