# A volume of saline solution weighed 25.6 g at 4 degrees C. An equal volume of water at the same temperature weighed 24.5 g. what is the density of the saline solution?

Mar 2, 2015

The density of the saline solution is 1.045 g/mL.

The density of pure water is 1.000 g/mL.

Since the volumes are the same, and the saline solution has more mass,

$d = \text{1.000 g/mL" × "25.6 g"/"24.5 g" = "1.045 g/mL}$

Mar 2, 2015

The density of the saline solution is $\text{1.04g/mL}$.

$\text{density}$ = $\text{mass"/"volume}$

Pure water is at its most dense at $\text{4"^("o")"C}$, which is $\text{1.000g/mL}$, which is equal to $\text{1.000g/cm"^3}$. At this point, the volume in mL or $\text{cm"^3}$ is equal to the mass of the water sample. So the volume of 24.5g of pure water is 24.5mL and $\text{24.5cm"^3}$.

$v o l u m {e}_{\text{water}}$ = $\left(m a s {s}_{\text{water"))/(density_("water}}\right)$ = (24.5"g")/(1.000"g"/("mL")) = 24.5mL of water.

Since the volume of the water is the same as the volume of the saline solution, we now know the mass and volume of the saline solution and can calculate its density.

$\mathrm{de} n s i t {y}_{\text{saline}}$ = $\left(m a s {s}_{\text{saline"))/(volume_("saline}}\right)$ = $\left(25.6 \text{g")/(24.5"mL}\right)$ = $\text{1.04g/mL}$