This would result in an increase of the calculated molarity than the actual one because if the droplets of titrant are present on the wall means there is actually less titrant used in the reaction with NaOH and some is left on the walls which would give readings more than it should and hence the calculated Molarity of NaOH would be more than it should actually as the equivalents consumed of the titrant would seem to be more than that are actually consumed by the NaOH.
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