A student performs a titration with different burettes and finds titre values of 25.2 mL, 25.25 mL and 25.0 mL. The number of significant figures in the average titre value

Question

Saurabh Koranglekar · Answer

Arun · Answer

We have 25.20,25.25 and 25.00 ml. Average= 3 25.20+25.25+25.00 ​ = 3 75.45 ​ = 25.15=25.1 In case of division, the final answer will contain as many significant figures as there are in an number with least significant numbers. So, the significant number is 3.

Krrish · Answer

we have 25.20, 25,25 and 25.0 mlavg.= (25.20+25.25+25.0)/3=75.5/3 (75.5 because when we add significant figures the result cannot have more digits to the right of the decimal point than either of the original numbers which in this case is 25.0. therefore we will have to round off 75.45 to 1 decimal place which is 75.5)=25.16666...=25.2(since when we divide significant figures, our answer cannot have more significant figures than either of the original numbers, ignoring exact numbers like 1,2,3..(basically natural numbers)).number of significant figures in 25.2 = 3Therefore our answer is 3

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A student performs a titration with different burettes and finds titre values of 25.2 mL, 25.25 mL and 25.0 mL. The number of significant figures in the average titre value

A student performs a titration with different burettes and finds titre values of 25.2 mL, 25.25 mL and 25.0 mL. The number of significant figures in the average titre value

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A student performs a titration with different burettes and finds titre values of 25.2 mL, 25.25 mL and 25.0 mL. The number of significant figures in the average titre value

A student performs a titration with different burettes and finds titre values of 25.2 mL, 25.25 mL and 25.0 mL. The number of significant figures in the average titre value

3 Answers

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