Question

If the CaCl2 was available not as a solid, but rather as a 0.40 molar solution, how many mL of that solution should be used if you desire to make 1.0 grams of MgCl 2?

A) 50 mL
B) 100 mL
C) 200 mL
D) 400 mL

Answer 1

To find out how many mL of a 0.40 M CaCl2 solution is needed to make 1 gram of MgCl2, we need to perform a stoichiometric calculation to calculate the moles of MgCl2 and then apply the molarity concept to find the volume in mL of the CaCl2 solution.

Step-by-step explanation:

The student is asking how many milliliters (mL) of a 0.40 molar (M) CaCl2 solution are needed to prepare 1.0 gram of MgCl2. To answer this, we will need to employ stoichiometry and molarity concepts. First, we must find the molar mass of MgCl2, which is 24.31 g/mol (Mg) + 2 × 35.45 g/mol (Cl) = 95.21 g/mol. Knowing that we need 1.0 gram, we can calculate the moles of MgCl2: (1.0 gram)/(95.21 g/mol) = 0.0105 moles of MgCl2. With the balanced chemical equation, we can deduce that 1 mole of CaCl2 will produce 1 mole of MgCl2, so 0.0105 moles of MgCl2 would also require 0.0105 moles of CaCl2.

Answer 2

To make 1.0 grams of MgCl₂ using a 0.40 molar solution of CaCl₂, you should use approximately 200 mL of that solution. (option C)

Step-by-step explanation:

First, determine the moles of MgCl₂ needed by dividing the given mass by its molar mass:

Moles of MgCl₂ = Mass / Molar Mass = 1.0 g / 95.211 g/mol

Next, consider the balanced chemical equation for the reaction between CaCl₂ and MgCl₂:

CaCl₂(aq) + MgCl₂(aq) → 2 CaCl₂(aq) + MgCl₂(aq)

This equation indicates that the stoichiometric ratio between CaCl₂ and MgCl₂ is 2:1. Therefore, the moles of CaCl₂ required are twice the moles of MgCl₂.

Moles of CaCl₂ = 2 × Moles of MgCl₂

Now, use the concentration of the CaCl₂ solution (0.40 mol/L) to find the volume (V) required:

`\[ V = \frac{{2 * \text{{Moles of MgCl₂}}}}{{\text{{Concentration of CaCl₂}}}} \]`

Substitute the values and convert the volume to milliliters:

`\[ V = \frac{{2 * \text{{Moles of MgCl₂}}}}{{0.40}} * 1000 \]`

This calculation gives the volume of the CaCl₂ solution needed to produce 1.0 gram of MgCl₂, which is approximately 200 mL. The subscript and superscript style is used here to represent the chemical formulas and stoichiometric coefficients without latex formatting.(option C)