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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