A 0.211 mol sample of an unknown gas contained in a 5.00 L flask is found to have a density of 2.50 g/L. The molecular weight of the unknown gas is g/mol.?

Md Ariful Islam

0 like 0 dislike
1 views
Share with your friends

1 Answers

Md Ariful Islam

Call

The idea here is that the of the gas tells you the mass of this gas that occupies exactly $"1 L"$ at some unspecified conditions for pressure and temperature.

In this case, you know that the gas had a density of $"2.50 g L"^(-1)$. This tells you that $"2.50 g"$ of this gas occupy exactly $"1 L"$ at the conditions for pressure and temperature used in the experiment.

You also know that the total volume of the flask is equal to $"5.00 L"$. At this point, you can use the density of the gas to calculate the mass of gas needed in order for the sample to occupy the given volume.

$5.00 color(red)(cancel(color(black)("L"))) * "2.50 g"/(1color(red)(cancel(color(black)("L")))) = "12.5 g"$

Now, in order to find the molar mass of the gas, you need to find the mass of exactly $1$ mole. You know that this sample contains $0.211$ moles of gas and has a mass of $"12.5 g"$, so you can say that $1$ mole will have a mass of

$1 color(red)(cancel(color(black)("mole gas"))) * "12.5 g"/(0.211 color(red)(cancel(color(black)("moles gas")))) = "59.24 g"$

Therefore, you can say that the molar mass of the gas is equal to

$color(darkgreen)(ul(color(black)("molar mass = 59.2 g mol"^(-1)))$

This means that $1$ mole of this gas has a mass of $"59.2 g"$.

The answer is rounded to three .

0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
A flask was weighed dry and then filled with water. If the masses of the empty flask and filled flask were 56.12 g and 87.39 g, respectively, and the density of water is 0.9976 g/$cm^3$, how would you calculate the volume of the flask in $cm^3$?

The idea here is that you can sue the mass of the empty flask and the mass of the filled flask to find the mass of water that can be...

1 Answers 1 views
A piece metal weighing 20.32 grams is added to a flask with a volume of 24.5 mL. It is found of that 18.52 grams of water (d = 0.9970 g/ml) be added to fill the flask. What is the density of the metal?

Find the volume of water added to fill the flask. $Density = (mass)/(volume)$ $Volume = (mass)/(density)$ $Vol (mL) = (18.52g)/((0.9970g)/(mL))$ $Vol (mL) = (18.52g)*((mL)/(0.9970g))$ $Vol (mL) = 18.58mL$...

1 Answers 1 views
1.25 g of an unknown gas at 69 °C and 1.05 atm is stored in a 1 L flask. What is the density of the gas?

So all you have to do is take a simple quotient. If we compressed the gas, how do you expect the to change?

1 Answers 1 views
A 0.176 mol sample of Kr gas is contained in a 8.00 L flask at room temperature and pressure. What is the density of the gas, in grams/liter, under these conditions?

First, we need to find the mass of $"Kr"$ by using the equation: $m/M_r=n$, where: $m$ = mass ($g$) $M_r$ = molar mass ($gcolor(white)(l)mol^-1$) $n$ = number of moles...

1 Answers 1 views
How many moles of $CO_2$ gas are contained in a flask holding 55.0 g of gas?

$"Number of moles"$ $=$ $(55.0*cancelg)/(44.01*cancelg*mol^-1)=1.25*mol$ Note that our mathematical expression was consistent dimensionally, the $g$ cancelled out, and $1/(mol^-1)=1/(1/(mol))=mol$ as required.

1 Answers 1 views
Asample of a compound contains 0.150 mol C, 0.105 mole H, 0.0300 mol O, and 0.0450 mol N. The molecular mass of the compound is found to be 402 g/mol. What is the compound's chemical formula?

Step 1. Calculate the empirical formula. The empirical formula is the simplest whole-number ratio of atoms in a compound. The ratio of atoms is the same as the ratio of...

1 Answers 1 views
A carbon compound contains 12.8% of carbon, 2.1% of hydrogen and 85.1% of bromine. The molecular weight of the compound is 187.9. What is the molecular formula of the compound? (Atomic weight: H = 1.008, C = 12.0, Br = 79.9)

As always with these problems, it is useful to assume $100*g$ of unknown compound....and thus find an empirical formula... $"Moles of carbon"=(12.8*g)/(12.011*g*mol^-1)=1.066*mol.$ $"Moles of hydrogen"=(2.1*g)/(1.00794*g*mol^-1)=2.084*mol.$ $"Moles of bromine"=(85.1*g)/(79.90*g*mol^-1)=1.066*mol.$ And we...

1 Answers 1 views
An unknown sample with a molecular mass of 180.0g is analyzed to yield 40% C, 6.7% H, and 53.3% O. What is the empirical formula and the molecular formula of this compound?

We assume $100*g$ of compound and work out the atomic proportions: $C: (40.0*g)/(12.011*g*mol^-1)$ $=$ $3.33$ $mol$. $H: (6.7*g)/(1.0794*g*mol^-1)$ $=$ $6.65$ $mol$. $O: (53.3.0*g)/(15.999*g*mol^-1)$ $=$ $3.31$ $mol$. We divide thru by...

1 Answers 1 views
An unknown compound was found to have a percent composition as follows: 47.0% potassium, 14.5% carbon, and 38.5% oxygen. What is empirical formula? If the true molar mass of the compound is 166.22 g/mol, what is its molecular formula?

As with all these problems, we assume a $100*g$ mass of unknown compound, and then we work out the molar quantity: $"Moles of potassium"=(47.0*g)/(39.10*g*mol^-1)=1.20*mol$ $"Moles of carbon"=(14.5*g)/(12.011*g*mol^-1)=1.21*mol$ $"Moles of oxygen"=(38.5*g)/(16.0*g*mol^-1)=2.41*mol$...

1 Answers 1 views
A sample of $KClO_3$ on decomposition gives 448 $mL$ of oxygen gas at NTP. Calculate: (a) Weight of oxygen produced (b) Weight of $KClO_3$ originally taken (c) Weight of $KCl$ produced?

Start with the balanced chemical equation for the of potassium chlorate, $KClO_3$ $color(green)(2)KClO_(3(s)) stackrel(color(red)("heat"))(->) color(blue)(2)KCl_((s)) + color(orange)(3)O_(2(g))$ The important thing to notice here is that $color(green)(2)$ moles of potassium chlorate...

1 Answers 1 views
X