Trial 12 1.Volume of Acid used ( mL) 2020 2.Molarity of acid used (M) 0.0160.016 3.Initial volume reading of base buret 018.6 4.Final volume reading of base used 18.636.9 5.Volume of base used (4-3, mL) 6.Molarity of base (1*2/5, M) 7.Average molarity of base

-- Volume of base used

--- Molarity of base

Explanation:

Given

-- Volume of acid used

--- Buret Initial reading

--- Buret Final reading

--- Molarity of the acid

Solving (a): Volume of base used (VB)

This is calculated by subtracting the initial reading from the final reading of the base buret.

i.e.

Solving (b): Molarity base (MB)

This is calculated using:

Make MB the subject

This gives:

Solving (c): There is no such thing as average molarity

the answer is true. hope it .

yes

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