Electrical conductivity of solution depends on the concentration of ions present and on the magnitude of charge present on the ions .
(a) 0.150M CaCl2 solution would have higher electrical conductivity . Because of higher charge on the Ca2+ ions .
(b)
Concentration of Co(NO3)2 = (14.22g/182.943g/mol)×(1000/57.8) = 1.3448M
Concentration of CuCl2 =( 9.528g/134.452g/mol)×(1000/32.9) = 2.154 M
The charge factor is same in both solutions, but number of ions in copper (II) chloride is more, hence 9.528g of copper (II) chloride in 32.9ml of water has higher electrical conductivity . (Answer)
predict which solution will have high electrical conductivity Question 2: Give one example of a situation with high...
Question 3: Predict which solutions will have a higher electrical conductivity. Justify your choice for full credit (4 pts) A. 0.200 M NaCl or 0.150 M CaCl2 B. 14.22 g of cobalt(II) nitrate in 57.8 mL of water or 9.528 g of copper (II) chloride in 32.9 mL of water.
Predict which solutions will have a higher electrical conductivity: 14.22 g of cobalt (II) nitrate in 57.8 mL of water or 9.528 g of copper (II) chloride in 32.9 mL of water.