Question

2.8 x10^-4kg of a gas occupies a volume of 224mL at 0°C and 1.013 x10^5Pa. What is the gas?
A.CO
B.CO2
C.Cl2
D. any gas

Answer 1

@maheshmeghwal9 , do you mind on working on this together?

Answer 2

no i won't; is this eudiometry?

Answer 3

or stoichiometry?

Answer 4

I have no clue lol, but it is in part of my practice book for my entrance exam :)

Answer 5

O.O
It must use the formula
\[PV=nRT \implies PV= {{m }\over {M}}RT.\]

Answer 6

so this is what I have so far:
0°C=273K, 1.013x10^5=1atm, V=0.224L and mass= 0.00000028g=2.8x10^-7g

Answer 7

okay then I should use the formula to solve for n? @maheshmeghwal9

Answer 8

no for "M"

Answer 9

okay so what is do m, and M stand for?

Answer 10

because n= moles

Answer 11

m= mass u have; i.e. - 0.00000028g=2.8x10^-7g
M = molar wt. of substance whose mass u have 0.00000028g=2.8x10^-7g.

Answer 12

okay thanks, i'll try that then :) so equation is M=mRT/PV

Answer 13

ya of course:)

Answer 14

Plz tell what value for "M" do u get?

Answer 15

when u do this:)

Answer 16

ok, i'll :)

Answer 17

:)

Answer 18

@maheshmeghwal9 i got 2.6 :/

Answer 19

so i think the answer is any gas
becoz the molar wt. we gt corresponds to \[H_2 = 2. \space \text{And we can take;} \space \space 2.8 \approx 2.\]

Answer 20

sorry
2.8 ->2.6*

Answer 21

idk, because the answer should be A

Answer 22

i gt A but from a different method:/

Answer 23

okay, how did you do it? i have to go grocery shopping so if you could just post your method that will be great! thx btw

Answer 24

k!
Since 0.224 l of gas contains 2.8 x 10^-4 kg
Therefore 22.4 l {I mean one mole} of gas will contain = 0.028 kg. = 28gm

& CO 's molar wt. is also 28gm
So Answer {A.}

Answer 25

but i don't know
why there is no need of pressure & temperature?
:/

Answer 26

'l' doesn't mean one mole there.
"i mean one mole" is there

Answer 27

@chmvijay Please help:)

Answer 28

the answer is CO

Answer 29

do u need how

Answer 30

okay, i kind of understand how you got it @maheshmeghwal9 , @chmvijay do you know how to get the answer?

Answer 31

mean i dint get u

Answer 32

do you know how to answer the question?

Answer 33

yaaa same thing what he has given already the equation just put the vaalues u get the molwecular weight u compare that with given gases molecular weight u will find it

Answer 34

@chmvijay My main problem is that i have found 2 answers from 2 different methods.

Answer 35

but which one is correct?

Answer 36

which one is other method
by doing that how much answer u got

Answer 37

tell me the answer first

Answer 38

the answer told by @bronzegoddess is option 'A'

Answer 39

yaaaa it is correct only the answer is CO

Answer 40

& my second solution is
Since 0.224 l of gas contains 2.8 x 10^-4 kg
Therefore 22.4 l {I mean one mole} of gas will contain = 0.028 kg. = 28gm
& CO 's molar wt. is also 28gm So Answer {A.}

but i don't know why there is no need of pressure & temperature in this method? :/

& my first solution was
PV =m/M RT.
but i gt option D.

Answer 41

why i m & @bronzegoddess getting option D
in
PV=m/M RT method?