Question

Strontium chloride hexahydrate $\left({\text{SrCl}}_{\text{2}}{\text{.6H}}_{\text{2}}\text{O}\right)$crystallizes in the trigonal system in a unit cell with $\mathit{a}\mathbf{=}\mathbf{89649}\stackrel{\mathbf{\circ }}{\mathbf{A}}$and $\mathit{\alpha }\mathbf{=}\mathbf{100}\mathbf{.}\mathbf{576}\mathbf{°}$. The unit cell contains three formula units. Compute the density of this substance.

Short answer

The density of this substance is $1.95g{\text{cm}}^{-3}$.

Step-by-step solution

The volume of this cell in cubic angstroms.

In trigonal systems,

$\alpha =\beta =\gamma \ne 90°$

$a=b=c$

_{$V=abc\sqrt{1-{\cos }^2\alpha -{\cos }^2\beta -{\cos }^2\gamma +2\cos \alpha \cos \beta \cos \gamma }$}

localid="1660647719041" $V=\left(8.9649\right)\left(8.9649\right)\left(8.9649\right)\sqrt{1-{\cos }^2\left(100.576°\right)-{\cos }^2\left(100.576°\right)-{\cos }^2\left(100.576°\right)+2\times \cos \left(100.576°\right)\cos \left(100.576°\right)\cos \left(100.576°\right)}$

$V=720.5\sqrt{1-\left(-0.18\right)\left(-0.18\right)-\left(-0.18\right)\left(-0.18\right)-\left(-0.18\right)\left(-0.18\right)+2\left(-0.18\right)\left(-0.18\right)\left(-0.18\right)}$

$V=720.5\sqrt{1-\left(0.0324\right)-\left(0.0324\right)-\left(0.0324\right)-0.011664}$

$V=720.5\sqrt{0.89116}$

$V=720.5\left(0.994\right)$

role="math" localid="1660647823315" $V=680.16\stackrel{\circ }{A^3}$

The density of this cell.

As;

$$\begin{gathered} density=\frac{mass}{volume} \\ density=\frac{z\times M}{V\times N_A} \end{gathered}$$

Here, the unit cell contains three formula units, i.e., z = 3.

And;

$\begin{array}{rcl}M\left(SrC{l}_2.6H_{2}O\right)& =& 158.53+108.12g/mol\\ & =& 266.65g/mol\end{array}$

Therefore;

$density=\frac{{\displaystyle 3\times 266.65g/mol}}{{\displaystyle 680.16\stackrel{\circ }{A^3}\times 6.022\times {10}^{23}/mol}}$

Changing Angstrom to cm;

$$\begin{gathered} 680.16\stackrel{\circ }{A^3}=680.16\times {\left({\text{10}}^{-8}\right)}^{3}c{m}^3 \\ 680.16\stackrel{\circ }{A^3}=680.16\times {\text{10}}^{-24}c{m}^3 \end{gathered}$$

So;

$density=\frac{3\times 266.65g/mol}{680.16\times {10}^{-24}c{m}^3\times 6.022\times {10}^{23}/mol}$

$$\begin{gathered} density=\frac{799.95}{409.59}g{\text{cm}}^{-3} \\ density=1.95g{\text{cm}}^{-3} \end{gathered}$$