Question

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

solution?​

Answer 1

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

Step-by-step explanation:

1. Calculate the moles of CaCl₂.

`\text{Moles} = \text{40.0 g} * \frac{\text{1 mol}}{\text{110.98 g}} = \text{0.3604 mol}`

Step 2. Calculate the volume of CaCl₂ solution

`\begin{array}{rcl} n &= &(c)/(V)\\\\\frac{\text{0.600 mol}}{\text{1 L}} &=& \frac{\text{0.3604 mol}}{V}\\\\ \frac{0.600V}{\text{1 L}} & = & 0.3604\\\\0.600V &= & \text{0.3604 L}\\V & = & \frac{\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 $\large \boxed{\textbf{601 mL}}$}`