Ask question

Ask question

All categories

andreyandreev [35.5K]

3 years ago

13

A sample of an unknown gas effuses in 14.4 min. An equal volume of H2 in the same apparatus under the same conditions effuses in

2.42 min. What is the molar mass of the unknown gas?

1 answer:

Elena-2011 [213]3 years ago

4 0

Answer:- molar mass of the unknown gas is 71.5 gram per mol.

Solution:- From Graham's law of effusion rates, the rate of effusion of a gas is inversely proportional to the square root of it's molar mass.

When we compare the effusion rates of two gases then the formula for Graham's law is:

$\frac{rate_1}{rate_2}=\sqrt{\frac{M_2}{M_1}}$

In this formula, V stands for volume and M stands for molar mass

Rate is volume effused per unit time. Since, the volumes are same, the formula could be written as:

$\frac{t_2}{t_1}=\sqrt{\frac{M_2}{M_1}}$

let's say in formula, subscript 1 is for hydrogen gas and 2 is for the unknown gas.

Molar mass of hydrogen is 2.02 grams per mol and the time taken to effuse it is 2.42 min. The time taken to effuse the unknown gas is 14.4 min and we are asked to calculate it's molar mass. let's plug in the values in the formula:

$\frac{14.4}{2.42}=\sqrt{\frac{M_2}{2.02}}$

$5.95=\sqrt{\frac{M_2}{2.02}}$

doing squares to both sides:

$35.4=\frac{M_2}{2.02}$

$M_2=35.4*2.02$

$M_2=71.5$

So, the molar mass of the unknown gas is 71.5 grams per mol.

You might be interested in

11. Assuming that the gases are ideal, calculate the amount of work done (joules) in each of the following reactions at 25 degre

gayaneshka [121]

<u>Answer:</u>

<u>For a:</u> Work done for the given reaction is 2477.572 J.

<u>For b:</u> Work done for the given reaction is 0 J

<u>Explanation:</u>

To calculate the work done for the reaction, we use the equation:

$W=-P\Delta V$

Ideal gas equation follows:

$PV=nRT$

Relating both the above equations, we get:

$W=-\Delta n_gRT$ ......(1)

where,

$\Delta n_g$ = difference in number of moles of products and reactants = $n_g_{(products)}-n_g_{(reactants)}$

R = Gas constant = 8.314 J/K.mol

T = temperature = $25^oC=[273+25]K=298K$

<u>For a:</u>

The chemical reaction follows:

$4HCl(g)+O_2(g)\rightarrow 2Cl_2(g)+2H_2O(g)$

$\Delta n_g=4-5=-1$

Putting values in equation 1, we get:

$W=-(-1mol)\times (8.314J/K.mol)\times 298K=2477.572J$

Hence, work done for the given reaction is 2477.572 J.

<u>For b:</u>

The chemical reaction follows:

$2NO(g)\rightarrow N_2(g)+O_2(g)$

$\Delta n_g=2-2=0$

Putting values in equation 1, we get:

$W=-(0mol)\times (8.314J/K.mol)\times 298K=0J$

Hence, work done for the given reaction is 0 J.

3 0

3 years ago

The chemical equation below shows the reaction between carbon dioxide (CO2) and lithium hydroxide (LiOH).

mariarad [96]

$M_{CO_{2}}=44,01\frac{g}{mol}\\
m=25,5g\\\\
n=\frac{m}{M}=\frac{25,5g}{44,01\frac{g}{mol}}\approx0,58mol$

CO₂ + 2LiOH ⇒ Li₂CO₃ + H₂O
1mol : 2mol
0,58mol : x

$x=\frac{0,58mol*2mol}{1mol}=\underline{1,16mol}$

6 0

3 years ago

Read 2 more answers

In a metal lattice, the metal ions are _____ and they are _____.

den301095 [7]

Answer:

a.)cations; free to move around

b.)cations; fixed in place

5 0

3 years ago

Rank the following gases in order of decreasing rate of effusion. Rank from the highest to lowest effusion rate. To rank items a

Alinara [238K]

Answer:

H2, He, PH3, F2, C5H10

Explanation:

From Graham's law, we understood that lighter gas will diffuse faster than heavier gas under same condition. Graham's law of diffusion states as follow:

The rate (R) of diffusion of a gas is inversely proportional to the square root of its density (D) provided temperature and pressure remains constant. Mathematically, it is represented as:

R & 1/√D

Recall:

Molar Mass (M) = 2 x vapour density (D)

M = 2D

R & 1/√M

From the above, we can see that the rate is inversely proportional to the square root of the molar mass of substance. This implies that lighter gas will diffuse faster.

Now, to rank the above from the highest to rate rate of effusion, let us determine the molecular weight of each substance. This is illustrated below:

Molar Mass of F2 = 19 x 2 = 38g/mol

Molar Mass of He = 4g/mol

Molar Mass of C5H10 = (12X5) + (10X1) = 70g/mol

Molar Mass of H2 = 2x1 = 2g/mol

Molar Mass of PH3 = 31 + (3x1) = 34g/mol

Now, we can rank the substance beginning from the highest to the lowest rate of effusion as follow:

Substance >> Molar Mass >> Rank

H2 >>>>>>>>> 2g/mol >>>>>> 1

He >>>>>>>>> 4g/mol >>>>>> 2

PH3 >>>>>>>> 34g/mol >>>>> 3

F2 >>>>>>>>>> 38g/mol >>>> 4

C5H10 >>>>>> 70g/mol >>>> 5

3 0

3 years ago

WHAT IS THE MOLAR MASS FOR AMMONIUM SULFIDE?<br> A. 59.1g/mol<br> B.82.2g/mol<br> C.50.1g/mol

Jobisdone [24]

Answer:

68g/mol

Explanation:

The formula of ammonium sulfide is:

Ammonium sulfide = (NH₄)₂S

The molar mass of a compound is the mass in gram of one mole of the substance. In a compound, it is expressed gram formula mass or gram-molecular weight.

It is determined by the addition of the component atomic masses and then expressed in grams;

Atomic mass of N = 14, H = 1 and S = 32

(NH₄)₂S = 2[14 + 4(1)] + 32 = 36 + 32 = 68g/mol

3 0

3 years ago