A volume of 250 ml of a 0.05 M solution of a reagent of formula weight (relative molecular mass) 40 was made up, the weighing being done by difference. The standard deviation of each weighing was 0.0001 g: what were the standard deviation and relative standard deviation of the weight of reagent used? The standard deviation of the volume of solvent used was 0.05 ml. Express this as a relative standard deviation. Hence calculate the relative standard deviation of the molarity of the solution
A volume of 250 ml of a 0.05 M solution of a reagent of formula weight (relative molecular mass) 40 was made up, the weighing being done by difference. The standard deviation of each weighing was 0.0001 g: what were the standard deviation and relative standard deviation of the weight of reagent used? The standard deviation of the volume of solvent used was 0.05 ml. Express this as a relative standard deviation. Hence calculate the relative standard deviation of the molarity of the solution
Chapter16: Applications Of Neutralization Titrations
Section: Chapter Questions
Problem 16.18QAP
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A volume of 250 ml of a 0.05 M solution of a reagent of formula weight (relative molecular mass) 40 was made up, the weighing being done by difference. The standard deviation of each weighing was 0.0001 g: what were the standard deviation and relative standard deviation of the weight of reagent used? The standard deviation of the volume of solvent used was 0.05 ml. Express this as a relative standard deviation. Hence calculate the relative standard deviation of the molarity of the solution. Repeat the calculation for a reagent of formula weight 392.
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