Chemistry Labs

Problem 4

A volume of 31.7 cm3^3 of a 0.1-normal NaOH solution is required for the neutralisation of 0.19 g of an organic acid whose vapour is thirty times as dense as gaseous hydrogen. Give the name and structural formula of the acid. (The acid concerned is a common organic acid.)
Step 2 of 3: Molar mass from titration
← PreviousNext →
M(acid)=0.19 g3.17×10−3/v mol=60 v g mol−1M(\text{acid}) = \dfrac{0.19\ \text{g}}{3.17 \times 10^{-3}/v\ \text{mol}} = 60\,v\ \text{g mol}^{-1}
Analysis

Dividing the weighed mass by the amount of acid gives M=0.19×v/3.17×10−3≈60 vM = 0.19 \times v / 3.17 \times 10^{-3} \approx 60\,v g mol−1^{-1}.