A cinema hall has equidistant rows 1 m apart. The length of the cinema hall is 287 m and it has 287 rows. From one side of the cinema hall, laughing gas $(N_{2} O)$ is released and from the other side, tear gas $(C_{6} H_{5} C O C H_{2} C l)$ is released. In which rows, spectators will be laughing and weeping simultaneously?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$187^{t h} row from N_{2} O and 100^{t h} row from C_{6} H_{5} C O C H_{2} C l$

$287^{t h} row from N_{2} O and 187^{t h} row from C_{6} H_{5} C O C H_{2} C l$

$100^{t h} row from C_{6} H_{5} C O C H_{2} C l and 287 from N_{2} O$

$100^{t h} row from both$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Molar weight of $C_{8} H_{7} O c l = 8 \times 12 + 7 \times 1 + 16 + 35.5 = 154.5 g$
Molar weight of $N_{2} O = 2 \times 14 + 16 = 44 g$
According to Graham’s law of diffusion,
$\frac{r N_{2} O}{r C_{8} H_{7} O C l} = \sqrt{\frac{M C_{8} H_{7} O C l}{M N_{2} O}} = \sqrt{\frac{154.5}{44}} = \sqrt{3.5} = 1.87 : 1$
$d N_{2} O = \frac{1.87}{2.87} \times 287 = 187^{th} row from N_{2} O side$
$d C_{8} H_{7} O C l = \frac{1.0}{2.87} \times 287 = 100^{th}$ row from weeping gas side
Therefore, the spectator from the side of $N_{2} O$ in the $187^{th}$ row will be laughing and weeping simultaneously. Alternatively, the spectator from the side of weeping as in the $100^{th}$ row will be laughing and weeping simultaneously.