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Question

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

Solution
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We have 25.20ml ,25.25ml and 25.00 ml.

Average= $$\frac{25.20+25.25+25.00}{3}$$
=$$\frac{75.45}{3}$$
=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.

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