Question

How many grams of potassium bromide, KBr, are in 100mL of a 0.50 M solution?

Answer 1

5.95g

Step-by-step explanation:

1
`dm^(3)` = 1000 mL

∴ 100 mL = 100 ÷ 1000 = 0.1
`dm^(3)`

Volume = 0.1
`dm^(3)`

Concentration = 0.5 M

Concentration =
`(No. of moles)/(volume)`

0.5 =
`(x)/(0.1)`

No. of moles = 0.5 x 0.1 = 0.05 moles

No. of moles =
`(mass)/(mass. in. 1. mole)`

Mass in 1 mole of KBr = 39 + 80 = 119g (39 is the mass of potassium and 80 is the mass of bromine)

0.05 =
`(x)/(119)`

x = 119 × 0.05 = 5.95g