Prelaboratory Assignment A 0. 2334-g quantity of primary standard NaCI is dissolved in 250.00 mL. A 10.00-mL aliquot of this sol Fajans titration. A chloride solu 250.00 mL and taking a 10-mL aliquot of the resulting solution, requires 11.21 mL of the same lution consumes 14.32 mL of a silver nitrate solution before reaching the end point in a tion, prepared by dissolving 0.2789 g of an unknown sample in silver nitrate solution to reach th contain? e end point. What percentage chloride (w/w) does the unknown
Solution
no of mol of NaCl dissolved = w/Mwt
= 0.2334/58.5
= 0.004 mol
concentration of NaCl solution = n/v
= 0.004/0.25
= 0.016 M
from concept of neutralisation
M1V1 = M2V2
M1 = molarity of NaCl taken = 0.016 M
V1 = volume of NaCl solution = 10 ml
M2 = molarity of AgNO3 = ?
v2 = Volume of AgNO3 consumed = 14.32 ml
(0.016*10) = (14.32*M2)
M2 = 0.0112 M
No of mol of AgNO3 required to react unknown = 11.21*0.0112
= 0.126 mmol
no of mol of chloride solution reacted = 0.126 mmol
mass of chloride present in sample = 0.126*10^-3*35.5
= 0.0045 g
%(w/w) of chloride present in sample = 0.0045/0.2789*100
= 1.61%
.
HMCS Vancouver Pre-Deployment Brief - May 2024 (Web Version).pptx
Prelaboratory Assignment A 0- 2334-g quantity of primary standard NaCI.docx
1. Prelaboratory Assignment A 0. 2334-g quantity of primary standard NaCI is dissolved in 250.00
mL. A 10.00-mL aliquot of this sol Fajans titration. A chloride solu 250.00 mL and taking a 10-
mL aliquot of the resulting solution, requires 11.21 mL of the same lution consumes 14.32 mL of
a silver nitrate solution before reaching the end point in a tion, prepared by dissolving 0.2789 g
of an unknown sample in silver nitrate solution to reach th contain? e end point. What percentage
chloride (w/w) does the unknown
Solution
no of mol of NaCl dissolved = w/Mwt
= 0.2334/58.5
= 0.004 mol
concentration of NaCl solution = n/v
= 0.004/0.25
= 0.016 M
from concept of neutralisation
M1V1 = M2V2
M1 = molarity of NaCl taken = 0.016 M
V1 = volume of NaCl solution = 10 ml
M2 = molarity of AgNO3 = ?
v2 = Volume of AgNO3 consumed = 14.32 ml
(0.016*10) = (14.32*M2)
M2 = 0.0112 M
2. No of mol of AgNO3 required to react unknown = 11.21*0.0112
= 0.126 mmol
no of mol of chloride solution reacted = 0.126 mmol
mass of chloride present in sample = 0.126*10^-3*35.5
= 0.0045 g
%(w/w) of chloride present in sample = 0.0045/0.2789*100
= 1.61%
