Question

0.56 g of ZnCl2 were added to 20,000 g of soil. calculate the following 1. Amount of Zn added in terms of (mg Zn)/(kg soil) 2. Amount of Zn added in terms of mmol (+) Zn 2+/kg soil 3. cmol(+) of added Zn2+/kg of soil 4. mg of added Zn2+/g of this soil

Answer 1

mass of ZnCl2 = 0.56 gm, 1gm= 1000mg, 0.56gm= 0.56*1000mg =560 mg

mass of soil= 20000gm, 1000gm= 1Kg, 20000gm= (20000/1000) kg = 20 kg

1. Zn added/ kg of soil= 560/20= 28 mg/kg , moles of ZnCl2= mass of ZnCl2/molar mass= 0.56/132= 0.0042 moles

1mole= 1000mmol, 0.0042 mol = 0.0042*1000mmol=4.2 mmol

ZnCl2---------->Zn+2+2Cl-, 1 mole of ZnCl2 gives 1 mole of Zn+2. Hence mmol of Zn+2= 4.2

Zn added in terms of mmol of Zn+2/ kg = 4.2/20=0.21mmol/kg, 1 mole= 100centimole, 0.0042 mole =0.0042*100 centimole= 0.042 hence cmol of Zn+2/kg = 0.42/20 =0.021cmol/kg , atomic weight of Zn= 65.38, mas of Zn+2= moles of Zn+2* atomic weight= 65.38*0.0042 gm of Zn+2= 0.274 gm , mass of Zn+2/kg = 0.274/20 gm of Zn+2/ kg =0.0137 Zn+2/kg

0.56 g of ZnCl2 were added to 20,000 g of soil. calculate the following 1. Amount of Zn added in terms of (mg Zn)/(kg soil) 2. Amount of Zn added in terms of mmol (+) Zn 2+/kg soil 3. cmol(+) of added Zn2+/kg of soil 4. mg of added Zn2+/g of this soil