Problem 4

A volume of 31.7 cm

^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 3 of 3: Molar mass from vapour density

← Previous Next →

Intuition

A relative density of 30 vs H2 means the molar mass is 30 times 2 g/mol, not 30 g/mol.

M(\text{acid}) = 30 \times M(\ce{H2}) = 30 \times 2 = 60\ \text{g mol}^{-1} \Rightarrow v = 1

Analysis

At equal T and p, densities are proportional to molar masses, so $M = 30 \times 2 = 60$ g mol $^{-1}$ . Comparison with the titration result $M = 60v$ gives $v = 1$ : a monoprotic acid of molar mass 60 g mol $^{-1}$ — acetic acid, $\ce{CH3COOH}$ .