Question 2 of 6: Heavy-Liquid Density Mixing (TBE/CCl 4 )

Question 2: Heavy-Liquid Density Mixing (TBE/CCl₄) (6 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $\rho_{\text{TBE}}=2.96\ \text{g/cm}^3$; $\rho_{\text{CCl}_4}=1.58\ \text{g/cm}^3$; target mixture density $\rho_{\text{mix}}=2.00\ \text{g/cm}^3$.

Find. The wt% TBE in the CCl₄/TBE mixture.

Approach. Assume ideal volume additivity on mixing (standard for this class of miscible heavy-liquid problem). On a 1 g basis, let $x$ = mass fraction TBE; the two component volumes must sum to the mixture's own volume, $1/\rho_{\text{mix}}$.

1. Set up the volume-additivity equation. $$\frac{x}{\rho_{\text{TBE}}}+\frac{1-x}{\rho_{\text{CCl}_4}}=\frac{1}{\rho_{\text{mix}}}$$ $$\frac{x}{2.96}+\frac{1-x}{1.58}=\frac{1}{2.00}=0.500$$
2. Solve for $x$. Multiplying through by $2.96\times1.58=4.677$: $$1.58x+2.96(1-x)=2.339 \quad\Rightarrow\quad -1.38x=-0.621 \quad\Rightarrow\quad x=0.4504$$ $$\boxed{\text{TBE concentration}=45.0\ \text{wt}\%}$$

Check: at $x=0.450$, $V=0.450/2.96+0.550/1.58=0.1522+0.3481=0.5003\ \text{cm}^3$/g, giving $\rho=1/0.5003=2.00\ \text{g/cm}^3$ – confirms the mix. In practice a laboratory prepares a small bracketing series of TBE/CCl₄ mixtures around 45 wt% (e.g. 40%, 45%, 50%) and titrates each against a small pure-mineral test grain, refining the blend until the grain hangs stationary in suspension – that stationary point is the direct experimental check that the calculated density has actually been achieved, since laboratory-grade densitometers are not always available at the bench.