# Prelaboratory Assignment A 0- 2334-g quantity of primary standard NaCI.docx

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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% .

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### 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%
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