Question

How many grams of Cabr2 would be needed to create 450 mL of a 2.00 M solution?

Answer 1

Mass = 179.9 g

Step-by-step explanation:

Given data:

Volume of solution = 450 mL

Molarity of solution = 2.00 M

Mass in gram required = ?

Solution:

Volume of solution = 450 mL× 1 L / 1000 mL = 0.45 L

Molarity = number of moles of solute/ Volume of solution in L

2.00 M = number of moles of solute / 0.45 L

Number of moles of solute = 2.00 M × 0.45 L

M = mol/L

number of moles of solute = 0.9 mol

Mass of CaBr₂ in gram:

Mass = number of moles × molar mass

Mass = 0.9 mol ×199.89 g/mol

Mass = 179.9 g