Answer:
The molarity (M) of the following solutions are :
A. M = 0.88 M
B. M = 0.76 M
Explanation:
A. Molarity (M) of 19.2 g of Al(OH)3 dissolved in water to make 280 mL of solution.
Molar mass of Al(OH)3 = Mass of Al + 3(mass of O + mass of H)
= 27 + 3(16 + 1)
= 27 + 3(17) = 27 + 51
= 78 g/mole
[tex]Al(OH)_3[/tex] = 78 g/mole
Given mass= 19.2 g/mole
[tex]Mole = \frac{Given\ mass}{Molar\ mass}[/tex]
[tex]Mole = \frac{19.2}{78}[/tex]
Moles = 0.246
[tex]Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}[/tex]
Volume = 280 mL = 0.280 L
[tex]Molarity = \frac{0.246}{0.280)}[/tex]
Molarity = 0.879 M
Molarity = 0.88 M
B .The molarity (M) of a 2.6 L solution made with 235.9 g of KBr
Molar mass of KBr = 119 g/mole
Given mass = 235.9 g
[tex]Mole = \frac{235.9}{119}[/tex]
Moles = 1.98
Volume = 2.6 L
[tex]Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}[/tex]
[tex]Molarity = \frac{1.98}{2.6)}[/tex]
Molarity = 0.762 M
Molarity = 0.76 M
