A student mixed 20.00mL of 0.100 M Na3PO4(aq) with 80.00 mL of 0.200 M Na2SO4 (a

Question

A student mixed 20.00mL of 0.100 M Na3PO4(aq) with 80.00 mL of 0.200 M Na2SO4 (aq) in a conical flask to make up solution A. She then transferred 10.00 mL solution A to a 50.00-mL volumetric flask and added distilled water up to the calibration mark to make up solution Calculate the molarity of the sodium ions in solution B.

(ANSWER: 0.0760 mol/L)

An unknown mass of potassium chromate is dissolved in distilled water to make 100.00 mL of stock solution. Then 25.00 mL of the stock solution is transferred to a 500.00-mL volumetric flask and enough distilled water is added up to the 500.00-mL mark to form a dilute solution. The concentration of the potassium ions in the dilute solution is 2.000 x 10^-3 mol/dm^3 Calculate the mass of potassium chromate dissolved to form the 100.00-mL stock solution

(ANSWER: 0.3884 g)

Explanation / Answer

moles of Na3PO4= 0.1*20/1000 = 0.002

Na3PO4--------> 3Na+ + PO4-3

1 moles of Na3PO4 gives 3 moles of Na+ ions

Moles of Na+ =3*0.002= 0.006

moles of Na2SO4= 0.2*80/1000=0.016

Na2SO4--------> 2Na+ SO4-2

Moles of Na+ =2*0.016= 0.032

Total moles of Na+=0.006+0.032=0.038 moles

total volume after mixing = 20+80 =100 ml =0.1L

100 L contains 0.038 moles

10ml contains 0.038*10/1000=0.0038 moles

Concentration = 0.0038*1000/50 = 0.076 mole/L

b) potasium chromate (K2CrO4)

let the mass be =m , molecular weight = 194

Moles of K2CrO4= m/194

100 ml contains m/194 moles

25ml contains (m/194*4)= m/776 moles of K2CrO4 ( 1 mole of K2CrO4 contains 2 moles ogf K+

Moles of K+ =2m/777

Concentration of K+when transferred to 500ml = m*(1000/500)/776 = 4m/776

4m/776= 2*10-3

m= 2*10-3*776/4 =0.387 gms

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