Question

What mass of calcium chloride is needed to prepare 2.657 L of a 1.56 M solution?

Answer 1

Answer:

459.126 grams of calcium chloride is needed to prepare 2.657 L of a 1.56 M solution

Explanation:

Molarity is a measure of the concentration of a solute in a solution that indicates the amount of moles of solute that appear dissolved in one liter of the mixture. In other words, molarity is the number of moles of solute that are dissolved in a given volume.

The Molarity of a solution is determined by the following expression:

$Molarity=\frac{number of moles of solute}{volume}$

Molarity is expressed in units $\frac{moles}{liter}$

In this case:

• Molarity: 1.56 M= 1.56 $\frac{moles}{liter}$
• Number of moles of calcium chlorine= ?
• Volume= 2.657 liters

Replacing:

$1.56 M=\frac{Number of moles of calcium chlorine}{2.657 liters}$

Solving:

Number of moles of calcium chlorine= 1.56 M* 2.657 liters

Number of moles of calcium chlorine= 4.14 moles

In other side, you know:

• Ca: 40 g/mole
• Cl: 35.45 g/mole

Then the molar mass of the calcium chloride CaCl₂ is:

CaCl₂= 40 g/mole + 2* 35.45 g/mole= 110.9 g/mole

Now it is possible to apply the following rule of three: if in 1 mole there is 110.9 g of CaCl₂, in 4.14 moles of the compound how much mass is there?

$mass=\frac{4.14 moles*110.9g}{1 mole}$

mass= 459.126 g

459.126 grams of calcium chloride is needed to prepare 2.657 L of a 1.56 M solution