Question

The relative density of mercury is 13.6. Find the ratio of the mass of mercury of a certain volume of that of water having ten times the volume of mercury.

Answer 1

Step 1: Given data:

$$\begin{gathered} \mathrm{Relative}\mathrm{density}\mathrm{of}\mathrm{mercury}=13.6 \\ \mathrm{Volume}\mathrm{of}\mathrm{water}\left({\mathrm{V}}_{\mathrm{W}}\right)=10\mathrm{x}\mathrm{Volume}\mathrm{of}\mathrm{mercury}\left({\mathrm{V}}_{\mathrm{M}}\right) \\ \mathrm{Density}\mathrm{of}\mathrm{water}\left({\mathrm{\rho }}_{\mathrm{W}}\right)={10}^3\raisebox{1ex}{$\mathrm{kg}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{m}}^3.$}\right. \end{gathered}$$

Step 2: Finding the ratio of the mass of the mercury to a certain volume of that of water:

$$\begin{gathered} \because RelativeDensityofMercury=\frac{Densityofmercury({\rho }_M)}{Densityofwater} \\ 13.6=\frac{{\rho }_M}{{10}^3} \\ {\rho }_M=13.6x{10}^3\raisebox{1ex}{$\mathrm{kg}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{m}}^3.$}\right. \end{gathered}$$

$$\begin{gathered} \because Density=\frac{Mass}{Volume} \\ \therefore Mass=DensityxVolume \\ \frac{Massofthemercury\left(m_M\right)}{Massofthewater\left(m_W\right)}=\frac{{\rho }_{M}x{V}_M}{{\rho }_{W}x{V}_W} \\ =\frac{13.6x{10}^{3}x{V}_M}{{10}^{3}x{V}_W} \\ =\frac{13.6}{10} \\ =13.6:10 \\ =34:25 \end{gathered}$$

Hence the desired ratio is 34:25.