A chemist pours 200 grams of a liquid (density = 2.5 gm/mL) into a cube beaker with 5 cm edges and a graduated cylinder with a base radius of 5 cm and a height of 5 cm.
What is fraction of the liquid that needs to be put in the graduated cylinder such that the liquid height is the same in the beaker
and graduated cylinder?
See your post above.
To find the fraction of the liquid that needs to be put in the graduated cylinder, we need to determine the volume of the liquid in both the beaker and the graduated cylinder.
First, let's calculate the volume of the liquid in the beaker:
Volume of the beaker = (side length of the cube)^3 = (5 cm)^3 = 125 cm^3
Next, let's calculate the volume of the liquid in the graduated cylinder:
Volume of the graduated cylinder = (base area of the cylinder) × (height of the cylinder)
Base area of the cylinder = π × (radius of the base)^2 = π × (5 cm)^2 = 25π cm^2
Volume of the graduated cylinder = (25π cm^2) × (5 cm) = 125π cm^3
Now, let's calculate the mass of the liquid in the beaker:
Mass = density × volume
Mass = 2.5 gm/mL × 200 mL = 500 grams
Since the density is in grams per milliliter, we need to convert the volume from cubic centimeters (cm^3) to milliliters (mL) since they are equivalent.
To find the fraction of the liquid that needs to be put in the graduated cylinder, we will divide the volume of the liquid in the graduated cylinder by the total volume of the liquid in the beaker and the graduated cylinder.
Fraction = Volume of the graduated cylinder / (Volume of the beaker + Volume of the graduated cylinder)
Fraction = (125π cm^3) / (125 cm^3 + 125π cm^3)
Simplifying further, we divide the numerator and denominator by 125 cm^3:
Fraction = π / (1 + π)
Therefore, the fraction of the liquid that needs to be put in the graduated cylinder such that the liquid height is the same in the beaker and graduated cylinder is approximately π / (1 + π).