Answer
2.206% Cl-
Work Step by Step
Cl- (aq) + AgNO3 (aq) ---> AgCl (s) + NO3- (aq)
moles AgNO3 = (0.2970 M) x (0.05363 L) = 0.01593 mol AgNO3
2
Finding the mass of Cl- in the seawater:
(0.01593 mol AgNO3) x (1 mol Cl- / 1 mol AgNO3) x (35.45 g Cl- / 1 mol Cl-)
= 0.5647 g Cl-
3
mass % = (grams Cl- / grams seawater) x 100%
grams Cl- = 0.5647 g
For the mass of seawater, d = m/V, so m = dV
d = 1.024 g/mL ; V = 25.00 mL
mass of seawater = dV = (1.024 g/mL) x (25.00 mL) = 25.60 g seawater
4
mass % = (0.5647 g Cl- / 25.60 g seawater) x 100% = 2.206% Cl-
