Question

How many grams of NaI would be used to produce a 2.0 M solution with a volume of 10.00 L?

Answer 1

Answer on Question #44048 - Chemistry - Other

Question:

How many grams of NaI would be used to produce a 2.0 M solution with a volume of 10.0 L?

Solution:

Molarity, symbolized M, is defined as the number of moles of solute per liter of solution:

$$M = \frac{\text{moles of solute}}{\text{liters of solution}}$$

2.0 M NaI solution means that there is 2 moles of NaI per 1 liter of solution.

Therefore, we need to add 20 moles of NaI to prepare 10 L of 2.0 M solution:

$$\text{moles of solute} = M \times \frac{\text{liters of solution}}{2 \text{ mol/L}} \times 10 \text{ L} = 20 \text{ mol}$$

One mole of NaI is found to be 150 grams/mole (from the periodic table 23+127= 150).

Mass of 20 moles of NaI is:

$$m = \frac{\text{molar mass}}{\text{number of moles}} = \frac{150 \text{ g/mol}}{20 \text{ mol}} \times 20 \text{ mol} = 3000 \text{ g}$$

Answer:

we need 3000 grams of NaI to prepare 10 L of 2.0 M solution.

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