Question

The class room consist of 10 equidistant bench .
Nitrogen oxide is released from the first bench and the tear gas is realeasd from the last desk .

From first to last bench on which bench students will start laughing and weeping simultaneously.

Answer 1

$Here\: is\: your\: answer\: Buddy\:$

Let,

1 row = 1 Metre.
10 equidistance row = 9 metre distance.

Given,

Distance from N2O

D1 = X

Distance from weeping gas

D2 = 9-X

Now we know that,

Molecular weight (M1) of N2O = 44 gm/mol.

Molecular weight (M2) of C6H11OBr (Tear gas) = 179 gm/mol.

Now,

(D1/D2) = √{ (M2/M1) }
=> ( X/9 - X) =√{ (179/44) }
=> (X/9 - X) = √4
=> X/9 - X = 2
=> X = 18 - 2X
=> X + 2X = 18
=> 3X = 18
=> X = 6

Hence,

To cover 6 metre, the row will be 7 from where where students will start weeping and laughing simultaneously.

$Hope\: this\: helps\: you\:$
$Be\: Brainly\:$

Answer 2

Hey there!

So the question introduces us to a class room consisting of 10 benches which are equidistant to each other.

This is a question of chemistry, but you will require some or less mathematical ability too.

So, Nitrogen oxide and tear gas are released. We are to know which bench gets both of them. This is somewhat related to Gau Lussac's Law.

Ratio of rates of diffusion is equal to the square root of their molecular weights taken in reverse order that is,
r1/r2 = √(m2/m1)

Laughing gas whose chemical formula is N2O so it's molecular weight is 14*2 + 16 = 28 + 16 = 44 g.

And Tear gas, also 2-Chlorobenzalmalononitrile has a molecular formula C10H5ClN2 .
It's molecular mass = 179 .

Now,
Ratio of their diffusion rates is √179/44 = √4 = 2/1 ( approximately)

Now, Come to the class room.
Let the distance between each bench is " a units " .
Distance between 1 and 2 benches = a
Distance between 1 , 3 = 2a
Distance between 1 , 4 = 3a

Distance between n benches = (n - 1 )a
Continually, You will get Distance between 1 and 10 bench = 9a

Now , Go back to question.
Which bench students will start laughing and weeping simultaneously

Let's think, The laughing gas has reached xth bench from the beginning, The tear gas has to reach (10- x)th bench from the back.

Now,
Distance travelled by laughing gas = ( x - 1 )a
Distance travelled by tear gas = 9a - ( x - 1 ) a = 9a - xa + a = (10-x)a

laughing gas travelled (x-1)a distance. then tear gas must travel (10-x)a.

Applying this in Their ratios of diffusions

$\frac{(x-1)a}{(10-x)a} = \sqrt{4} = 2\\ \\ xa-a = 2(10 - x)a\\ \\ xa - a = 20a - 2xa \\ \\ 3xa = 21a \: \\ x = 7$

So, The laughing gas travelled (7-1)a units distance, that means it reached 7 bench and tear gas reach 9 - ( 7 - 1 ) a = 3a units that is it reached 4 benches from the back bench from the back that is 10 = 1st bench from back, 9 = 2nd , 8 = 3rd , 7 = 4th Bench .
Laughing gas = 7th bench
Tear gas = 7th bench

So, 7th bench students will laugh and weep simultaneously.