A student prepares 150. mL of a 2.00 M Ba(OH)2 aqueous solution in lab. What mass of Ba(OH)2 is fully dissolved in the solution?

51.4g

Explanation:

Given that:

Amount of moles of Ba(OH)2 (n) = ?

Volume of Ba(OH)2 solution (v) = 150.0mL

[Convert 150.0mL to liters

If 1000 mL = 1L

150.0mL = 150.0/1000 = 0.150L]

Concentration of Ba(OH)2 solution (c) = 2.00M

Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence

c = n / v

make n the subject formula

n = c x v

n = 2.00M x 0.150L

n = 0.3 mole

Now given that,

Amount of moles of Ba(OH)2 (n) = 0.3

Mass of Ba(OH)2 in grams (m) = ?

For molar mass of Ba(OH)2, use the molar masses:

Barium, Ba = 137.3g;

Oxygen, O = 16g;

Hydrogen, H = 1g

Ba(OH)2 = 137.3g + [(16g + 1g) x 2]

= 137.3g + [17g x 2]

= 137.3g + 34g

= 171.3 g/mol

Since, amount of moles = mass in grams / molar mass

0.3 mole= m / 171.3g/mol

m = 0.3 mole x 171.3g/mol

m = 51.39g

[Round the 51.39g to the nearest tenth, so it becomes 51.4g]

Thus, the mass of Ba(OH)2 fully dissolved in the solution is 51.4 grams

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