Question

How many grams of CaBr2 would be needed to create 450. mL of a 2.00 M

solution? Round off your answer to the nearest gram. Record your answer in the
grid below.​

Answer 1

Mass of CaBr₂ = 179.9 g

Step-by-step explanation:

Given data:

Volume of CaBr₂ solution = 450.0 mL (450/1000 = 0.45 L)

Molarity of solution = 2.00 M

Mass of CaBr₂ = ?

Solution:

Formula:

Molarity = number of moles / volume of solution in L

Now we will put the values.

2.00 M = number of moles / 0.45 L

Number of moles = 2 M × 0.45 L

Number of moles = 0.9 mol ( M = mol/L)

Mass of CaBr₂:

Mass of CaBr₂ = number of moles × molar mass

Mass of CaBr₂ = 0.9 mol × 199.89 g/mol

Mass of CaBr₂ = 179.9 g