Question #98930

1 Answer
Feb 5, 2016

Answer:

#"770. g"#

Explanation:

The key to this problem lies with the specific gravity of blood.

For a given liquid, its specific gravity is defined as the ratio between its density and the density of a reference substance, usually water at #4^@"C"#.

As an interesting fact, the reference substance is water at #4^@"C"# because that's the temperature at which water's density is maximum #-># more on that in this answer.

So, you know that

#color(blue)("S.G." = rho_"liquid"/rho_"water")#

Now, water's density is never actually equal to #"1 g/mL"#, but you can approximate it to be equal to #"1 g/mL"#, especially at #4^@"C"#.

So, if you taje #rho_"water" = "1 g/mL"#, you can find the density of blood by using its specific gravity

#"S.G." = rho_"blood"/rho_"water" implies rho_"blood" = "S.G." xx rho_"water"#

#rho_"blood" = 1.05 * "1 g/mL" = "1.05 g/mL"#

All you have to do now is convert the volume of blood from pints to milliliters, since that is the unit used in the density of blood.

The conversion factor here is

#"1 US pint " = " 473.176 mL"#

This means that the volume of blood will be equivalent to

#1.55 color(red)(cancel(color(black)("pints"))) * "473.176 mL"/(1color(red)(cancel(color(black)("pint")))) = "733.42 mL"#

You know that blood has a density of #"1.05 g/mL"#. This means that every milliliter of blood has a mass of #"1.05 g"#.

In your case, the #"733.42-mL"# sample will have a mass of

#733.42 color(red)(cancel(color(black)("mL"))) * overbrace("1.05 g"/(1color(red)(cancel(color(black)("mL")))))^(color(purple)("density of blood")) = color(green)("770. g")#

The answer is rounded to three sig figs.