An analytical chemist weighs out 0.126 g of an unknown monoprotic acid into a 250 mL volumetric flask and dilutes to the mark with distilled water. He then titrates this solution with 0.1400 MNaOH solution. When the titration reaches the equivalence point, the chemist finds he has added 15.0 mL of NaOH solution. Calculate the molar mass of the unknown acid. Be sure your answer has the correct number of significant digits. \( \square \frac{\mathrm{g}}{\mathrm{mol}} \)
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- Volume of NaOH = 15.0 mL = 0.0150 L - Molarity of NaOH = 0.1400 M \[ \text{Moles of NaOH} = \text{Molarity} \times \text{Volume} = 0.1400 \, \text{mol/L} \times 0.0150 \, \text{L} = 0.00210 \, \text{mol} \] Show more…
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