Question

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

Answer 1

Answer:

5.95g

Explanation:

1 $dm^{3}$ = 1000 mL

∴ 100 mL = 100 ÷ 1000 = 0.1 $dm^{3}$

Volume = 0.1 $dm^{3}$

Concentration = 0.5 M

Concentration = $\frac{No. of moles}{volume}$

0.5 = $\frac{x}{0.1}$

No. of moles = 0.5 x 0.1 = 0.05 moles

No. of moles = $\frac{mass}{mass. in. 1. mole}$

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

0.05 = $\frac{x}{119}$

x = 119 × 0.05 = 5.95g