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3 years ago

Are Molar Mass and Molecular Mass the same thing? If they are different, then why are they used interchangeably?

2 answers:

7 0

The molar mass of one mole of whatever substance/element you have while the molecular mass is the mass of one molecule. They are used differently according to the measurements and base on their units. I hope you are satisfied with my answer.

exis [7]3 years ago

4 0

At this point there are two answers: one says that there is no difference, except for the units; the molecular weight (MW) is the mass of one molecule, while a molar mass is the mass of a mole (6.02 x 10 ^ 23) of molecules.

While in certain contexts both statements can be shown to be true, I'm not sure these answers really provide you with the insight you're looking for.

So, we'll look at your question from a different perspective. Take a real example, something simple like water. Water has a nominal MW of 18 (so we'll use nominal precision for simplicity, instead of the 4-places often used for these types of calculations).

The MW of water (formula = H2O) is the weight of one atom of oxygen, which = 16 amu [8-neutrons at 1 amu each plus 8 protons at 1 amu each = 16 amu), plus two atoms of hydrogen, at 1 proton (1 amu) each. Normal everyday hydrogen has no neutrons. So for H2O, we have a total molecular weight of 18 amu.

From the CRC reference book we find that one amu weighs 1.66 x 10 ^--24 grams. Multiplying the two and in keeping with the two units of precision we're working with, one molecule of water has a mass of 29.8 x 10^-24g, or [3.0 x 10 ^ -23 g] per molecule of water.

A Mole is simply Avagadro's number (6.02 x 10 ^ 23) of anything... protons, baseballs, whatever. The term Molar Mass in chemistry refers to the mass of a mole of molecules. So in this case a molar mass of water molecules is Avagadro's number of them, the mass therefore being [6.02 x 10 ^ 23] x 3.0 x 10 ^ -23 g/ molecule] = 18.0 g

Summarizing:

Molecular Weight is the weight given in amu of an atom or molecule. For H2O, the MW is 18 amu or 3.0 x 10 ^ -23 g.

Molar Weight is the weight, usually in grams of 6.02 x 10 ^ 23 measurements which happens to be equal to the MW of the molecule (or atomic wt. of the atom) , and for water is 18.0 g.

So, while MW and molar weight are related, their absolute values are magnitudes apart.

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lana [24]

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3 years ago

Which gas law (choices are Charles' Law, Gay-Lussac's Law, or Boyle's Law) explains each scenario:________.

posledela

Answer:

A. Boyle's Law

B. Charles' Law

C. Gay-Lussac's Law

Explanation:

An air bag inflates due to the decomposition of sodium azide or NaN₃ to completely fill the bag with nitrogen gas which is an example of Boyle's law, which states that the pressure of a given mass of gas is inversely proportional to its volume, hence due to the estricted volume of the airbag, the pressure of the nitrogen gas in the bag increses protecting the occupants of a cr from injuries in a crash

Helium balloon decrease in sice in a freezer is an example of Charlles law which states that the volume of a given mass of gas is nverslely proportionl to its temperature at constant pressure

A can of spray paint will explode if tossed into a fire is an example of Gay-Lussac's Law which states that the pressure of a given mass of gas is directly proportional to its temperature hence the increased pressure causes the can ti explode

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3 years ago

What is the answer to 4x - 16

andrezito [222]

It simply equals 4x-16.......
the expression is in lowest terms, so you can’t do anything else

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3 years ago

Read 2 more answers

determine the freezing point depression of a solution that contains 30.7 g glycerin (c3h8o3, molar mass

Blababa [14]

The freezing point depression of a solution containing 30.7 g of glycerin is calculated as -1.65°C

Equating :

It is given that,

Given mass of glycerin is = 30.7 grams (Solute)

Volume of water = 376 mL

$K_{f}$ or molar -freezing-depression point is = 1.86°C/m

Molar mass of glycerin = 92.09 g/mole

Now, to work out the value, the mass of water should be known. Thus, to calculate, the formula used will be:

Mass = Density X Volume

Mass = 1.0 g/mL X 376 mL

Mass = 376 g or 0.376 Kg

Using the formula of melting point depression, the equation becomes:

Δ $T_{f}$ = i × $K_{f}$ ×m

T⁰- $T_{s}$ = $i *K_{f} *\frac{mass of glycerin}{molar mass of glycerin * mass of water in kg}$

in which,

Δ $T_{f}$ = change in freezing point

Δ $T_{s}$ = freezing point of solution that has to be find

ΔT° = freezing point of water ()

Since, glycerin is a non-electrolyte, the Van't Hoff factor will be 1.

Substituting the values in the above equation:

0⁰C₋T $_{s}$ = 1 ×1.86°C/m × $\frac{30.7}{92.09g/mol * 0.376kg}$

$T_{s}$ = -1.65°C

Thus, the freezing point depression of a solution is -1.65°C

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Freezing point depression

Freezing point depression is a colligative property observed in solutions that results from the introduction of solute molecules to a solvent. The freezing points of solutions are all less than that of the pure solvent and is directly proportional to the molality of the solute

Is melting point elevation or depression?

Boiling point elevation is that the raising of a solvent's boiling point due to the addition of a solute. Similarly, melting point depression is the lowering of a solvent's freezing point due to the addition of a solute. In fact, because the boiling point of a solvent increases, its melting point decreases

Learn more about freezing point depression :

brainly.com/question/26525184

#SPJ4

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1 year ago

Suppose you work at a theme park. Your supervisor wants you to make a sign displaying the maximum weight that a roller coaster t

grin007 [14]

Answer:

amusement parks. Each day, we flock by the millions to the nearest park, paying a sizable hunk of money to wait in long lines for a short 60-second ride on our favorite roller coaster. The thought prompts one to consider what is it about a roller coaster ride that provides such widespread excitement among so many of us and such dreadful fear in the rest? Is our excitement about coasters due to their high speeds? Absolutely not! In fact, it would be foolish to spend so much time and money to ride a selection of roller coasters if it were for reasons of speed. It is more than likely that most of us sustain higher speeds on our ride along the interstate highway on the way to the amusement park than we do once we enter the park. The thrill of roller coasters is not due to their speed, but rather due to their accelerations and to the feelings of weightlessness and weightiness that they produce. Roller coasters thrill us because of their ability to accelerate us downward one moment and upwards the next; leftwards one moment and rightwards the next. Roller coasters are about acceleration; that's what makes them thrilling. And in this part of Lesson 2, we will focus on the centripetal acceleration experienced by riders within the circular-shaped sections of a roller coaster track. These sections include the clothoid loops (that we will approximate as a circle), the sharp 180-degree banked turns, and the small dips and hills found along otherwise straight sections of the track.

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2 years ago