chemisry

A 0.125 gram sample if a monoprotic acid of unknown molar mass is dissloved in water and titrated with 0.1003 MNaOH. The endpoint is reached after adding 20.77 ml of base. What is the molar mass of the unknown acid?

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  1. for the unknown
    moles = mass/RMM

    we have the mass, but we need the number of moles to find RMM

    so number of moles of NaOH used

    is 0.1003 x 20.77/1000

    this is the same as the number of moles of the unknown as it is a monoprotic acid.

    substitute the number of moles and the mass (0.125 g) in the equation to find RMM

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