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Chapter 4: Problem 53

Calculate the mass of \(\mathrm{KI}\) in grams required to prepare \(5.00 \times 10^{2} \mathrm{~mL}\) of a \(2.80 \mathrm{M}\) solution.

Short Answer

Expert verified
The mass of KI required to prepare the solution is 232.4 grams.

Step by step solution

01

Convert Volume to Liters

The volume is given in milliliters. We need to convert this to liters, since the volume in the formula for molarity is in liters. Recall that 1 Liter is equivalent to 1000 milliliters. Therefore, \(500 mL = 0.500 L\).
02

Calculate the Number of Moles Needed

Using the formula \(M = \frac {n}{V}\), we can determine the number of moles of KI required. Given the molarity of the solution is 2.80 M and we've already converted the volume to 0.500 L, we rearrange the formula \(n = M \times V\) to get \(n= 2.80 \times 0.500 = 1.4 moles\). Therefore, we need 1.4 moles of KI to prepare the solution.
03

Convert Moles to Grams

We have found that the number of moles of KI needed is 1.4 moles. Now we use the molar mass of KI to convert this to grams. Recall that the molar mass of KI is about 166 g/mol. Therefore, the mass of KI required can be calculated as \(mass = n \times molar~mass = 1.4 \times 166 = 232.4 grams\).

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