## Chemistry: Molecular Approach (4th Edition)

$\underline{\text{3}\text{.5 cm}}$
(1) Density (d) is defined as mass per unit volume. $d=m/V$ Here, m is the mass of mercury and V is the volume of mercury. Rearrange this equation to calculate the volume of mercury. $V=m\text{ }/d$ (1) (2) Calculate the volume of mercury at $\text{0}\text{.0}{}^\circ \text{C}$ by substituting the values in equation (1) as follows: \begin{align} & V\ \text{(0}\text{.0}{}^\circ \text{C})=m\text{/}d \\ & =\frac{3.380\text{ }g}{13.596\ \text{g/c}{{\text{m}}^{\text{3}}}} \\ & \text{= 0}\text{.2486 }c{{m}^{3}} \end{align} (3) Calculate the volume of mercury at $\text{25}\text{.0}{}^\circ \text{C}$ by substituting the values in equation (1) as follows: \begin{align} & V\text{ (25}\text{.0}{}^\circ \text{C})=m\text{/}d \\ & =\frac{3.380\text{ }g}{13.534\ \text{g/c}{{m}^{3}}} \\ & \text{= 0}\text{.2497 }c{{m}^{3}} \end{align} (4) Calculate the change in the volume of mercury as follows: \begin{align} & Change\text{ }in\text{ }volume=V\text{ (25}\text{.0}{}^\circ \text{C})-V\text{ (0}\text{.0}{}^\circ \text{C}) \\ & =\text{0}\text{.2497 }c{{m}^{3}}-\text{0}\text{.2486 }c{{m}^{3}} \\ & =\text{0}\text{.0011 }c{{m}^{3}} \end{align} (5) Capillary is cylindrical in shape. Let the rise in the capillary be $x\text{ cm}$. Calculate the change in volume in the capillary with the temperature as follows: \begin{align} & \text{Change in volume = 3}\text{.14}\times {{\text{(Radius of capillary)}}^{2}}\times \text{Rise in the capillary} \\ & \text{=3}\text{.14}\times {{\left\{ \left( \frac{0.200\text{ mm}}{2} \right)\left( \frac{1\text{ cm}}{10\text{ mm}} \right) \right\}}^{2}}\times \left( x\text{ cm} \right) \\ & =\text{0}\text{.000314}x\text{ c}{{\text{m}}^{3}} \end{align} (6) Calculate the rise in the capillary by equating the two-volume changes as follows: \begin{align} & \text{0}\text{.000314}x\text{ c}{{\text{m}}^{3}}\text{ = 0}\text{.0011 c}{{\text{m}}^{\text{3}}} \\ & x=\text{ 3}\text{.5 cm} \end{align} Rise in the mercury level in thermometer is $\underline{\text{3}\text{.5 cm}}$.