Answer:
5.95g
Explanation:
1 [tex]dm^{3}[/tex] = 1000 mL
∴ 100 mL = 100 ÷ 1000 = 0.1 [tex]dm^{3}[/tex]
Volume = 0.1 [tex]dm^{3}[/tex]
Concentration = 0.5 M
Concentration = [tex]\frac{No. of moles}{volume}[/tex]
0.5 = [tex]\frac{x}{0.1}[/tex]
No. of moles = 0.5 x 0.1 = 0.05 moles
No. of moles = [tex]\frac{mass}{mass. in. 1. mole}[/tex]
Mass in 1 mole of KBr = 39 + 80 = 119g (39 is the mass of potassium and 80 is the mass of bromine)
0.05 = [tex]\frac{x}{119}[/tex]
x = 119 × 0.05 = 5.95g
