Question

carbon tetrachloride, ccl4, was once a popular solvent in organic chemistry, but due of its adverse health effects (disrupting the central nervous system and causing liver and kidney damage) it is only used when absolutely necessary in a well ventilated area. how much energy is required to vaporize 57.7 ml of liquid carbon tetrachloride? at its melting point of -22.92°c, the heat of fusion, δhfus, is +2.52 kj mol-1 and at the normal boiling point of 76.72°c, the heat of vaporization, δhvap, is +32.54 kj mol-1 and the density is 1.59 g ml-1. report your answer in kilojoules to 4 sig figs, but do not include units in the answer.

Explanation:

Step1: Calculate the mass of \(CCl_4\)

Use the formula \(m=
ho V\), where \(
ho = 1.59\space g/mL\) and \(V = 57.7\space mL\).
\(m=1.59\times57.7 = 91.743\space g\)

Step2: Calculate the molar mass of \(CCl_4\)

The molar mass of \(CCl_4\): \(M=(12.01)+4\times(35.45)=153.81\space g/mol\)

Step3: Calculate the number of moles of \(CCl_4\)

Use the formula \(n=\frac{m}{M}\), where \(m = 91.743\space g\) and \(M = 153.81\space g/mol\)
\(n=\frac{91.743}{153.81}\approx0.5965\space mol\)

Step4: Calculate the energy required for vaporization

Use the formula \(q = n\Delta H_{vap}\), where \(n = 0.5965\space mol\) and \(\Delta H_{vap}=+ 32.54\space kJ/mol\)
\(q=0.5965\times32.54\approx19.41\)

Answer:

19.41