Question

a student was asked to solve the following problem: the density of liquid oxygen is 1.141 g/cm³. what volume (in ml) is present in a 18.4 - gram sample? review the picture below showing a template to answer the question using dimensional analysis. match the \block numbers\ to \answer choices\ to correctly setup the problem using dimensional analysis. block 1 block 2 block 3 answer i: $\frac{1 cm^{3}}{1 ml}$ answer ii: 18.4 g answer iii: $\frac{1 ml}{1 cm^{3}}$ answer iv: $\frac{1}{18.4 g}$ answer v: $\frac{1 cm^{3}}{1.141 g}$ answer vi: $\frac{1.141 g}{1 cm^{3}}$ block 3 block 1 block 2 multiple choice 10 points evaluate: 20 + 5×2+6

Explanation:

Step1: Recall density - volume - mass formula

The formula for density is $
ho=\frac{m}{V}$, where $
ho$ is density, $m$ is mass and $V$ is volume. We can re - arrange it to $V = \frac{m}{
ho}$. Here, $m = 18.4$ g and $
ho=1.141$ g/cm³.

Step2: Set up dimensional analysis

We know that $1$ cm³=$1$ mL. We want to cancel out the 'g' (gram) unit and get the volume in mL. The correct setup using dimensional analysis is $V=18.4\ g\times\frac{1\ cm^{3}}{1.141\ g}\times\frac{1\ mL}{1\ cm^{3}}$. So, Block 1 should be Answer II (18.4 g), Block 2 should be Answer V ($\frac{1\ cm^{3}}{1.141\ g}$) and Block 3 should be Answer III ($\frac{1\ mL}{1\ cm^{3}}$).

Answer:

Block 1: Answer II: 18.4 g
Block 2: Answer V: $\frac{1\ cm^{3}}{1.141\ g}$
Block 3: Answer III: $\frac{1\ mL}{1\ cm^{3}}$