3.6698 g of Na2CrO4 (MW = 161.97 g/mol) is dissolved in 100.0 mL of water. Assum

Question

3.6698 g of Na2CrO4 (MW = 161.97 g/mol) is dissolved in 100.0 mL of water. Assuming the solution has a density of 1.00 g/mL, what is the concentration of Na (MW = 22.9898 g/mol) in the solution in units of

Sapling Learning macmilan leami 3.6698 g of NapCrO4 (MW = 161.97 g/mol) is dissolved in 100.0 mL of water. Assuming the solution has a density of 1.00 g/mL, what is the concentration of Na (MW-22.9898 g/mol) in the solution in units of a) molarity (M)? Number b) parts per thousand (ppt)? Number ppt c) 10.0 mL of the solution is then diluted to a final volume of 1000.0 mL. What is the concentration of Nat in the diluted solution in units of parts per million (ppm)? Number ppm Previous Give Up & View Solution O Check Answer Next Exit Hint

Explanation / Answer

a)

Number of moles of Na2CrO4 = (3.6698 g/ 161.97 gmol-1) = 0.02265 moles

Number of moles of Na+ = 0.0265 x 2 = 0.0453 moles

Volume = 0.100L

Molarity = moles/L = (0.0453/0.100) = 0.453M

b)

161.97 g of Na2CrO4 contains 45.9796 g of sodium

3.6698 g of of Na2CrO4 contains : 1.0417 g of sodium

1.0417 g of sodium is present in 100 mL

1 ppt = one part in 1000.

The strength of solution in ppt = 1.0417 x 10 = 10.417 ppt

c)

V1 = 10 mL

M1 = 0.453 M

V2 = 1000 mL

M2 = ?

using M1V1 = M2V2

M2 = (M1V1)/V2 = (0.453 x 10)/1000 = 0.00453 M

So the concentration of diluted solution is 0.00453 M = 0.00453 mol L-1

The amount of Na+ ions present in the dilute solution= 0.00453 molL-1 x 45.9796 gmol-1

= 0.2083 g/L= 208.3 mg/L

1 ppm = 1 mg/L

so 208.3 mg/L = 208.3 ppm

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