Problem 102P

Acetone (fingernail-polish remover) has a density of 0.7857 g/cm3.

(a) What is the mass in grams of 17.56 mL of acetone?

(b) What is the volume in milliliters of 7.22 g of acetone?

Solution 102P

Here, we are going to find the mass of 17.56 mL of acetone. Also, we are going to calculate the volume in millilitres of 7.22 g of acetone.

Step1:

Density of any substance may be defined as mass per unit volume. If ‘d’ be the density, ‘m’ be the mass and ‘v’ be the volume of any substance, then, numerically,

d = ------- (1)

Step2:

(a) Given, volume of acetone = 17.56 mL = 17.56 cm3 [1 mL = 1 cm3]

Density of acetone = 0.7857 g/cm3

Substituting the values in equation (1), we get,

0.7857 g/cm3 = m / 17.56 cm3

m = 0.7857 x 17.56 [(g/cm3) x cm3]

m = 13.80 g

Thus, mass of 17.56 mL of acetone is 13.80 g.

Step3:

(b) Given, mass of acetone = 7.22 g

Density of acetone = 0.7857 g/cm3

Substituting the values in equation (1), we get,

0.7857 g/cm3 = 7.22 g / v

v = 7.22 / 0.7857 [g / (g/cm3)]

v = 9.19 cm3.

v = 9.19 mL [1 cm3 = 1 mL] Thus, volume of 7.22 g of acetone is 9.19 mL.