Final answer:
The total ion concentration in a 0.5 L hot pack solution with 0.162 grams of CaCl2 is 0.0088 M when rounded to the correct number of significant figures.
Step-by-step explanation:
To find the total ion concentration in a 0.5 L solution of CaCl2, we first need to calculate the number of moles of CaCl2. The molar mass of CaCl2 is 110.98 g/mol, and it's given that we have 0.162 grams of CaCl2. Using the relationship moles = mass / molar mass, we get:
moles of CaCl2 = 0.162 g / 110.98 g/mol = 0.00146 moles
Since each unit of CaCl2 dissociates into three ions (one Ca2+ and two Cl- ions), the total number of moles of ions is 3 times the moles of CaCl2, which is 0.00438 moles. To calculate the concentration, divide by the volume of the solution:
total ion concentration = 0.00438 moles / 0.5 L
= 0.00876 M
Thus, the total ion concentration in the hot pack solution is 0.00876 M, which should be reported as 0.0088 M when rounded to the correct number of significant figures.