Question

What is the molarity (M) of the following solutions?

Answer 1

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

$Al(OH)_3$ = 78 g/mole

Given mass= 19.2 g/mole

$Mole = \frac{Given\ mass}{Molar\ mass}$

$Mole = \frac{19.2}{78}$

Moles = 0.246

$Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}$

Volume = 280 mL = 0.280 L

$Molarity = \frac{0.246}{0.280)}$

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

$Mole = \frac{235.9}{119}$

Moles = 1.98

Volume = 2.6 L

$Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}$

$Molarity = \frac{1.98}{2.6)}$

Molarity = 0.762 M

Molarity = 0.76 M