Question

How many grams of cabr2 are needed to prepare 4.65 l of a 7.65 m solution?

Answer 1

Molarity is defined as the number of moles of solute dissolved in 1 L of solution 
the molarity of the solution to be prepared is 7.65 M
the volume to be prepared of CaBr₂ is 4.65 L
the number of CaBr₂ moles are - 7.65 mol/L x 4.65 L = 35.6 mol
the molar mass of CaBr₂ - 200 g/mol
mass of 35.6 mol - 35.6 mol x 200 g/mol = 7120 g
mass required of CaBr₂ to make the solution is 7120 g

Answer 2

The number of grams of CaBr₂ is 7.12 g.

What is mass?

Mass is the quantity of matter of a physical body.

Given, the molarity of the solution is 7.65 m

The volume to be prepared is  4.65 L

Calculating the moles

7.65 mol/L × 4.65 L = 35.6 mol

The molar mass of CaBr₂ is 200 g/mol

Now, the mass will be

35.6 mol × 200 g/mol = 7.12 g

Thus, the mass required is 7.12 g.

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