Question

PLEASE HELP!!!
A 19.4 gram sample of gas is found to have a volume of 1.3 liters at 282 K and 1.2 atm. What is the molar mass of this gas? Show all of the work used to solve this problem.

Answer 1

Use PV=nRT

Answer 2

here T=282
V=1.82
P=1.2
R=0.0821
and n=m/M we need M, so put m=19.4 to get M

Answer 3

It should look1.2*1.82=(19.4/M)*0.0821*282

Answer 4

@fenispart Got it?

Answer 5

@arindameducationusc Yeah I understand now. I got 289.55

Answer 6

@arindameducationusc I also need help with one last one:

A solution is made by dissolving 23.5 grams of glucose (C6H12O6) in .245 kilograms of water. If the molal freezing point constant for water (Kf) is -1.86 °C/m, what is the resulting Δtf of the solution? Show all the steps taken to solve this problem.

Answer 7

Didn't I give you the formula last time?
Okay I will show again...

Answer 8

\[\Delta{T}=K _{f}* W _{b}*1000/ Mw _{b}*W _{a}\]
Here Mwb= 180 because C6H12O6
6*12+12*1+6*1=180

Answer 9

convert .245 kg to gram....

Answer 10

25.4g

Answer 11

you get 245 gram
(This is Wa)

Answer 12

1kg=1000grams

Answer 13

and Wb is 19.4, and the rest is given in question...
Got it?

Answer 14

@fenispart Can you solve now?

Answer 15

Is 19.4 always Wb?

Answer 16

No not always, but for this question 19.4 is Wb

Answer 17

How do you find Wb?

Answer 18

hmmmm....Its given in the question.... READ THE QUESTION

Answer 19

Wait are we talking about the first question or the second one?

Answer 20

oops... my bad...... its second question and Wb is 23.5.

Answer 21

sorry about that

Answer 22

23.5 is given... I guess now you got it. Both questions are about gas so.... I made a mistake........

Answer 23

Can you do now? @fenispart

Answer 24

How do I find Kf? @arindameducationusc

Answer 25

Kf is given -1.86

Answer 26

I got -0.991 @arindameducationusc

Answer 27

Total answer?

Answer 28

Yes

Answer 29

@arindameducationusc

Answer 30

i m here

Answer 31

Yes your answer is correct....

Answer 32

I just checked it... Good job! Bravo!

Answer 33

Thank you so much!