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Standard XII

Chemistry

Millikan's Oil Drop Experiment

Assume that a...

Question

Assume that a box full of electrons had a total charge of $- 96368 C$ . The number of moles of eletcrons present inside the box is
$Given : e = - 1.6 \times 10^{- 19} C$

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Solution

Charge on an electron $= - 1.6 \times 10^{- 19} C$
Total charge conatined $= - 96368 C$
Number of electrons $= \frac{magnitude of total charge}{Magnitude of charge on one elctron}$

$= \frac{96368 C}{1.6 \times 10^{- 19} C} = 6.023 \times 10^{23}$

So, the box has 1 mol of electrons.

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Millikan's Oil Drop Experiment

Standard XII Chemistry

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Assume that a box full of electrons had a total charge of −96368 C. The number of moles of eletcrons present inside the box is Given:e=−1.6×10−19 C

Charge on an electron =−1.6×10−19 C Total charge conatined =−96368 C Number of electrons =magnitude of total chargeMagnitude of charge on one elctron =96368 C1.6×10−19 C=6.023×1023 So, the box has 1 mol of electrons.

Assume that a box full of electrons had a total charge of $- 96368 C$ . The number of moles of eletcrons present inside the box is
$Given : e = - 1.6 \times 10^{- 19} C$

Charge on an electron $= - 1.6 \times 10^{- 19} C$
Total charge conatined $= - 96368 C$
Number of electrons $= \frac{magnitude of total charge}{Magnitude of charge on one elctron}$

$= \frac{96368 C}{1.6 \times 10^{- 19} C} = 6.023 \times 10^{23}$

So, the box has 1 mol of electrons.