Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A sample of a liquid solvent has density \(0.915 \mathrm{g} / \mathrm{mL}\) What is the mass of \(85.5 \mathrm{mL}\) of the liquid?

Short Answer

Expert verified
The mass of 85.5 mL of the liquid solvent with a density of \(0.915 \frac{\text{g}}{\text{mL}}\) is approximately 78.1325 g.

Step by step solution

01

Write down the formula for density

The formula for density is given by: Density = Mass / Volume
02

Rearrange the formula to solve for mass

We will rearrange the formula to get the mass as the subject: Mass = Density × Volume
03

Substitute the given values

Now we will substitute the given values for density and volume into the formula: Mass = 0.915 g/mL × 85.5 mL
04

Calculate the mass

Now we will multiply the density and volume to get the mass of the liquid solvent: Mass = 0.915 g/mL × 85.5 mL = 78.1325 g (Note: since g/mL × mL gives grams) So, the mass of 85.5 mL of the liquid solvent is 78.1325 g.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Density Formula
Understanding the density formula is key to calculating the mass or volume of a substance when one of these properties is known. Density is a measure of how much mass is contained in a given unit volume of a material. It's usually expressed in grams per cubic centimeter (g/cm³) or grams per milliliter (g/mL) for liquids and solids, and kilograms per cubic meter (kg/m³) for gases.

The formula for density \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \) is a straightforward expression connecting these two physical properties. When the problem asks for mass, as with our liquid solvent, the formula can be rearranged to \( \text{Mass} = \text{Density} \times \text{Volume} \) which allows us to solve for mass if we know the substance's density and the volume it occupies. It's important to always use consistent units when performing calculations to ensure accuracy.
Mass and Volume
Mass and volume are intrinsic physical properties of substances. Mass represents the amount of matter in an object and is usually measured in grams (g) or kilograms (kg), whereas volume is the amount of space that matter occupies and can be measured in milliliters (mL), cubic centimeters (cm³), or liters (L). Our understanding of these properties comes into play when we calculate the density of a material, or as in the given problem, when we have the density and need to find the mass.

In the exercise, the volume of the liquid is known (85.5 mL), and with the density formula, these values are used to find the mass. It is crucial that the units for mass and volume correspond to each other, as is the case with grams and milliliters, to avoid conversion errors.
Unit Conversion
Unit conversion is necessary in many scientific calculations to ensure that measurements are in the right units for a particular equation or comparison. In density calculations, we often need to convert volumes from milliliters to cubic centimeters or grams to kilograms, especially if our density is given in a different unit than our volume or mass.

One key thing to remember is that 1 mL is equal to 1 cm³, which simplifies conversions for liquids. When dealing with more complex conversions, such as those between imperial and metric units, using a conversion factor or a conversion chart can be very helpful. Understanding how to convert units is essential in ensuring that the calculations are correct, and it also allows us to compare densities of materials regardless of the units originally used.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.

Sign-up for free