Question

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.

Answer 1

Solution:

The conductivity of the solution increases with increasing concentration.

A) 0.200 M NaCl and 0.150 M CaCl2

0.200 M NaCl has more electrical conductivity because of greater concencentration of Na+ and Cl- ions.

B) Copper (II) chloride is more conductive

Explanation:

Molarity of Cobalt(II) nitrate

= mass / (molar mass x volume in L)

= 14.22 g / 182.943 g mol-1 x 0.0578 L

= 1 34 M

Molarity of copper (II) chloride

= mass / molar mass x volume in L

= 9.528 g / 134.45 g mol-1 x 0.0329 L

= 2.15 M

Thus,

Copper (II) chloride is more conductive.