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Example 1.9: Computing Quantities from Measurement Results and Known Mathematical Relations

What is the density of common antifreeze in units of g/mL? A 4.00-qt sample of the antifreeze weighs 9.26 lb.

Solution

Since density\( = \dfrac{\mathrm{mass}}{\mathrm{volume}}\), we need to divide the mass in grams by the volume in milliliters. In general: the number of units of \(B = \mathrm{the}\) number of units of A × unit conversion factor. The necessary conversion factors are given in Table 6:

\(\mathrm{lb}\) \(= 453.59\ \mathrm{g}\)



\(\mathrm{qt}\) \(= \dfrac{1}{1.0567} \cdot 1\ \mathrm{L}\)

\(\ \ \ =0.946342 \cdot 1\ \mathrm{L}\)

\(\ \ \ =0.94634\ \mathrm{L}\)


\(V_{\mathrm{sample}}\) \(= 4.00 \cdot \mathrm{qt}\)

\(\ \ \ =4.00 \cdot 0.94634\ \mathrm{L}\)

\(\ \ \ =3.785\ \mathrm{L}\)


\(m_{\mathrm{sample}}\) \(= 9.26 \cdot \mathrm{lb}\)

\(\ \ \ =9.26 \cdot 453.59\ \mathrm{g}\)

\(\ \ \ =4200.\ \mathrm{g}\)


\(ρ_{\mathrm{sample}}\) \(= \dfrac{m_{\mathrm{sample}}}{V_{\mathrm{sample}}}\)

\(\ \ \ =\dfrac{4200.\ \mathrm{g}}{3.785\ \mathrm{L}}\)

\(\ \ \ =1110.\ \frac{\mathrm{g}}{\mathrm{L}}\)


\(ρ_{\mathrm{sample}}\) \(= 1110.\ \frac{\mathrm{g}}{\mathrm{L}}\)

\(\ \ \ =1.110\ \frac{\mathrm{g}}{\mathrm{mL}}\)