A stalagmometer gives 36 drops for a fixed volume of water at 20 °C, where γ = 72.8 mN/m and ρ =…

Step 1. The surface tension. $\gamma_x = \gamma_w \times \dfrac{n_w}{n_x} \times \dfrac{\rho_x}{\rho_w} = 72.8 \times \dfrac{36}{80} \times \dfrac{877}{998} = 28.8\ \text{mN/m}$.

Step 2. The rise ratio. In the same tube, $\dfrac{h_x}{h_w} = \dfrac{\gamma_x/\rho_x}{\gamma_w/\rho_w} = \dfrac{28.8/877}{72.8/998} = 0.450$.

Step 3. The rise. $h_x = 0.450 \times 29.7\ \text{mm} = 13.4\ \text{mm}$.

Step 4. The link. The ratio of the rises is $0.450$, the same as $\dfrac{n_w}{n_x} = \dfrac{36}{80}$. This is no accident. Both $h$ and $n$ depend on the ratio $\gamma/\rho$, and the drop count is inversely proportional to it.