This problem considers the random errors involved in making up a standard soluti

Question

This problem considers the random errors involved in making up a standard 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. Repeat the calculation for a reagent of formula weight 392.

Explanation / Answer

Part-A

Relative molecular mass of the reagent = 40

250 mL of 0.05M solution would have,

0.05 M = x mol/250 mL

xg = 0.05 x 0.25 = 0.0125 mol

1 mol of reagent = 40 g

therefore, 0.0125 mol would have,

0.0125 x 40 = 0.5 g

weighing being done by difference means the weight of the reagent is the diffrence between the weight of vial before and after adding the reagent.

Standard deviation of each weighing is 0.0001g

therefore, the standard deviation of the weight of reagent also would be 0.0001g

relative standard deviation(RSD) would be, standard deviation/mean value = 0.0001/0.5 = 0.0002.

% RSD = 0.0002/100 = 0.02 %.

b) standard deviation of the volume of solvent = 0.05 mL

relative standard deviation(RSD) would be, standard deviation/mean value = 0.05/250 = 0.0002.

% RSD = 0.0002/100 = 0.02 %.

c) Molarity of solution is moles/L (which in turn g/L).

Hence, relative standard deviation of the molarity of solution would be square root of (0.0002)^2+(0.0002)^2 = 0.0002828 (%RSD = 0.028 %)

Part-B

Relative molecular mass of the reagent = 392

250 mL of 0.05M solution would have,

0.05 M = x mol/250 mL

xg = 0.05 x 0.25 = 0.0125 mol

1 mol of reagent = 392 g

therefore, 0.0125 mol would have,

0.0125 x 392 = 4.9 g

Standard deviation of each weighing is 0.0001g

therefore, the standard deviation of the weight of reagent also would be 0.0001g

relative standard deviation(RSD) would be, standard deviation/mean value = 0.0001/4.9 = 0.00002040816.

% RSD = 0.002040816 %.

b) standard deviation of the volume of solvent = 0.05 mL

Relative standard deviation(RSD) would be, standard deviation/mean value = 0.05/250 = 0.0002.

% RSD = 0.0002/100 = 0.02 %.

c) Molarity of solution is moles/L (which in turn g/L)

Hence, relative standard deviation of the molarity of solution would be square root of (0.00002)^2+(0.0002)^2 = square root of 4 x 10^-10 + 4 x 10^-8 = 0.00020 (%RSD = 0.02 %)

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