Question

How many kilograms of water are in a 0.813 M strontium bromide, SrBr₂, solution that contains 3.18 g of strontium bromide?

Answer 1

Molar mass of SrBr2,

MM = 1*MM(Sr) + 2*MM(Br)

= 1*87.62 + 2*79.9

= 247.42 g/mol

mass(SrBr2)= 3.18 g

use:

number of mol of SrBr2,

n = mass of SrBr2/molar mass of SrBr2

=(3.18 g)/(2.474*10^2 g/mol)

= 1.285*10^-2 mol

use:

M = number of mol / volume in L

0.813 = 1.285*10^-2/ volume in L

volume = 0.01581 L

volume = 15.81 mL

Since density of solution 1 g/mL,

mass of solution = 15.81 g

mass of water = mass of solution - mass(SrBr2)

= 15.81 g - 3.18 g

= 12.6 g

= 0.0126 Kg

Answer: 0.0126 Kg