Question

3) Describe how to prepare the following solutions using solid sodium iodide, liquid water and common laboratory equipment (balance, spatulas, volumetric flasks, pipets, beakers, etc.) a) 200mL of 0.16M Nal b) 200mL of 0.16m Nal c) 200mL of 0.16% mass Nal

Answer 1

Answer:

1. To prepare : 200.0 mL of 0.16 M NaI solution.

Let us calculate firrst # of moles of NaI.

Molarity = 0.16 M, Volume of solution = 200.0 mL = 0.200 L

# of moles of NaI = Molarity * Volume in L = 0.16 * 0.200 = 0.032 mol.

Molar mass of NaI = 149.89 g.mol^-1.

Mass of NaI = Molar mass of NaI * # of moles of NaI = 0.032 mol * 149.89 g.mol^-1 = 4.796 g

4.796 g of NaI in 200.0 mL gives 0.16 M NaI solution.

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b) 200.0 mL of 0.16m NaI

4.796 g of NaI dissolved and diluted to 200.0 mL of distilled water.

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c) 200.0 mL of 0.16% NaI

A % solution is prepared by adding A gram of compound dissolved and diluted to 100 g of solvent.

Similarly,

0.16 % NaI is prepared by adding 0.16*2 = 0.32 g of NaI dissolved and diluted to 200 g of water gives 0.16 % NaI solution.

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