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Q.

The number of molecule at NTP in 1 ml of an ideal gas will be

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a

$2 .69 \times 10^{19}$

b

$2 .69 \times 10^{23}$

c

$6 \times 10^{23}$

d

None of these

answer is B.

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Detailed Solution

$∵$ 22400 ml at NTP has 6.023x 10²³ molecule
$∴$ 1 ml at NTP has = $\frac{6.023 \times 10^{23}}{22400}$
= 0.0002688x 10²³ = 2.69x 10¹⁹.

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The number of molecule at NTP in 1 ml of an ideal gas will be