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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)A. `(100pi)/(36 xx 22.4)`B. `(10pi)/(18 xx 22.4)`C. `(10pi)/(72 xx 22.4)`D. `(10pi)/(60 xx 22.4)` |
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Answer» Correct Answer - A End correction `= (0.3) d = 1 cm` `d = (10)/(3) cm` vol. of tube `= (pi(d^(2))/(4))l = (pi)/(4)((10)/(3))^(2) xx 100 cm^(2)` (take `I = 0.99 m =n 1 m`) `= (10pi)/(36) lit` moles `= (10 pi)/(36 xx 22.4)` moles (`22.4 "lt"`. contains `1` mole `(10pi)/(36)"lt"` contains `(10pi)/(36 xx 22.4)` mole) |
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