Answer: 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
