Question

part a
using the relationship between mass and volume in density, determine which conversion factor is needed to solve for the missing value where x indicates a numerical value equi
drag the appropriate labels to their respective targets.
view available hint(s)
given value need conversion factor
50 g ml of sample (\frac{1 ml}{x g})
55 ml g of sample (\frac{x g}{1 ml})
45 l kg of sample (\frac{x kg}{1 l})
70 kg l of sample (\frac{1 l}{x kg})

Explanation:

Step1: Recall density formula

Density $
ho=\frac{m}{V}$, where $m$ is mass and $V$ is volume. To convert from mass to volume, we use $\frac{V}{m}$, and from volume to mass we use $\frac{m}{V}$.

Step2: Analyze first - row

Given 50 g (mass) and need mL (volume) of sample. Since $
ho=\frac{m}{V}$, then $V = \frac{m}{
ho}$, so the conversion factor is $\frac{1\ mL}{X\ g}$.

Step3: Analyze second - row

Given 55 mL (volume) and need g (mass) of sample. Since $
ho=\frac{m}{V}$, then $m=
ho V$, so the conversion factor is $\frac{X\ g}{1\ mL}$.

Step4: Analyze third - row

Given 45 L (volume) and need kg (mass) of sample. Since $
ho=\frac{m}{V}$, then $m =
ho V$, so the conversion factor is $\frac{X\ kg}{1\ L}$.

Step5: Analyze fourth - row

Given 70 kg (mass) and need L (volume) of sample. Since $
ho=\frac{m}{V}$, then $V=\frac{m}{
ho}$, so the conversion factor is $\frac{1\ L}{X\ kg}$.

Answer:

Given Value	Need	Conversion Factor
55 mL	g of sample	$\frac{X\ g}{1\ mL}$
45 L	kg of sample	$\frac{X\ kg}{1\ L}$
70 kg	L of sample	$\frac{1\ L}{X\ kg}$