Answer to Question #144095 in General Chemistry for erwin james villanueva

Question

A dental amalgam has a mass of 1.005 kg and a volume of 433 mL. What is the density of the dental amalgam in lb per ft3?

Answer 1

Q144095

A dental amalgam has a mass of 1.005 kg and a volume of 433 mL. What is the density of the dental amalgam in lb per ft3?

Solution:

We have to find the density in lb / ft ³ units, but we are given the mass in ‘kg’ and volume in ‘mL’ .

So we will have to convert mass to lb, and volume to ft³ units.

Step 1: Convert 1.005 kg to ‘lb’ and 433 mL to ‘ft³ ‘ .

For converting kg --> lb, we can use the conversion factor, 1 kg = 2.2046 lb.

mass of dental amalgam in ‘lb’ = 1.005 kg * 2.2046 lb /1 kg = 2.2156 lb ;

For converting mL to ft³, we will have to use three conversion factors.

1cm³ = 1mL, 1inch = 2.54 cm ; and 1 ft = 12 inch ;

inch, cm and feet, all of them are units of length. We are finding volume, so a cube must be given to inch, cm and ft.

Volume of dental amalgam in ‘ft ‘ =

$=433mL*[1cm^3/1mL]*[1inch/2.54cm]^3 *[1ft/12inch]^3$ ;

= $433mL∗[1cm 3 /1mL]∗[1inch^3/2.54^3cm^3] ∗[1ft^3/12^3inch^3]$

= $433mL∗[1cm3/1mL]∗[1inch ^3 /16.387 cm ^3 ]∗[1ft ^3 /1728 inch ^3 ]$

$=433ft^3/(16.387*1728)$

$=433ft^3/28317$

= 0.01529 ft³ ;

Step 2 : Find the density in lb/ ft³ using the, mass = 2.2156 lb, and volume = 0.01529 ft³

We know the formula, Density = mass / Volume ;

substitute mass = 2.2156 lb and volume = 0.01529 ft³ in this formula, we have

Density = 2.2156 lb / 0.01529 ft³ = 144.9 lb/ft³ ;

In question the quantity with least number of significant figure is 433 mL, so our final answer must also be in 3 significant figure.

144.9 lb/ft³ in 3 significant figure is 145 lb/ft³ .

So the density of dental amalgam is 145 lb/ft³ .