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?
Step 1 of 3
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.
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)
(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
Textbook: Introductory Chemistry
Author: Nivaldo J Tro
Since the solution to 102P from 2 chapter was answered, more than 598 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Introductory Chemistry, edition: 5. The full step-by-step solution to problem: 102P from chapter: 2 was answered by , our top Chemistry solution expert on 05/06/17, 06:45PM. Introductory Chemistry was written by and is associated to the ISBN: 9780321910295. The answer to “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?” is broken down into a number of easy to follow steps, and 31 words. This full solution covers the following key subjects: acetone, mass, fingernail, grams, Density. This expansive textbook survival guide covers 19 chapters, and 2045 solutions.