Assume end correction approximately equals to `(0.3) xx` (diameter of tube), estimate the approximate number of moles of ir present inside the tube (Assume tube is at `NTP`, and at `NTP, 22.4` litre contains `1` mole)

Options

A. `(10 pi)/(36 xx 22.4)`
B. `(10 pi)/(18 xx 22.4)`
C. `(10 pi)/(72 xx 22.4)`
D. `(10 pi) /(60 xx 22.4)`

Answer

A

Explanation

End correction `= (0.3) d = 1 cm`
`d = (10)/(3) cm`
Volume of the tube
`= (pi (d^(2))/(4)) t = (pi)/(4) ((10)/(3))^(2) xx 100 cm^(3)`
(take `1 = 0.99 m ~~ 1 m)`
` = (10 pi)/( 36) litres`
moles `= (10 pi)/(36 xx 22.4) nodes`
(`22.4 l` contains `1 mol` and `10 pi//36` l contains `10 pi //36 xx 22.4 mol`)