20 g of CaCl2 is added to a small amount of water in the flask. Additional water

ID: 939043 • Letter: 2

Question

20 g of CaCl2 is added to a small amount of water in the flask. Additional water is added to bring the total volume of solution to 0.75 L.

(a) Calculate the concentration of calcium chloride in units of mg/L.

(b) Calculate the concentration of calcium chloride in units of Molarity.

(d) Calculate the concentration of calcium chloride in units of mg/L as CaCO3.

(e) 5 g of NaCl is added to the above CaCl2 solution. Calculate the total concentration (mg/L) of chloride in the solution.

a) mass of CaCl2 =20 g =20*103 mg, volume =0.75L

Concentration = mass/ volume in Litres =20*103/0.75=26667 mg/L

b) Molarity = moles of CaCl2/ Litre of solution=

Moles= mass/ Molecular weight of CaC2

Molecular weight of CaCl2= 40+2*35.5= 40+71=111

Moles =20/111=0.18018 moles

Molarity= 0.18018/0.75L=0.24M

c) Normality =gm equivalens/ liters

for Calcium chloride, equivalent weight= molecular weigh/2

Normality =20/(111/2)/0.75L =0.48N

d) moles of CaCl2=0.18018

Equivalent moles of CaCO3= 0.18018

mass of CaCo3= 0.18018*100 ( Molecular weight of CaCO3=100)= 18.018 gms

concentration = 18.018*1000 mg/0.75L=24.024*1000 mg/L

e) moles of CaCl2= 0.18018 CaCl2 ----> Ca+2 +2Cl-

1 mole of CaCl2 produces 2 moles of Cl-

0.18018 moles produces 2*0.18018=0.36036 moles of Cl-

Mole of Cl- produced by 5 gm of NaCl= 5/58.5 ( NaCl---> Na+ + Cl-)=0.08547

total Cl- = 0.36036+0.08547=0.44853 moles

Molecular weight of Cl = 35.5

Mass of Cl- = 0.44853*35.5=15.83 gm =15.83*1000 mg

Concentration in mg?l= 15.83*1000/0.75= 21107 mg/L

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