Question

What volume of solution would be present if 40.0 g Caci, would be used to make a 0.600M

Answer 1

Answer:

$\large \boxed{\text{601 mL}}$

Explanation:

1. Calculate the moles of CaCl₂.

$\text{Moles} = \text{40.0 g} \times \dfrac{\text{1 mol}}{\text{110.98 g}} = \text{0.3604 mol}$

Step 2. Calculate the volume of CaCl₂  solution

$\begin{array}{rcl} n &= &\dfrac{c}{V}\\\\\dfrac{\text{0.600 mol}}{\text{1 L}} &=& \dfrac{\text{0.3604 mol}}{V}\\\\ \dfrac{0.600V}{\text{1 L}} & = & 0.3604\\\\0.600V &= & \text{0.3604 L}\\V & = & \dfrac{\text{0.3604 L}}{\text{0.600 L}}\\\\ & = & \text{0.601 L}\\& = & \textbf{601 mL}\\\end{array}\\\text{The volume of the solution would be \large \boxed{\textbf{601 mL}} }$