A gas is present in a 30 litre vessel at 27o C and a pressure of 8-2 - 10-10 atmosphere- The number of molecules in the gas are-

The correct option is A 6.02× 10^14 n=PV/RT Number of molecules = n× N_o n=8.2 × 10^ 10× 30/0.082 × 300 nbsp;=10^ 9 Number of molecules ...

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Byju's Answer

Standard XII

Chemistry

Liquefaction of Gases and Critical Constants

A gas is pres...

Question

A gas is present in a 30 litre vessel at $27^{o} C$ and a pressure of $8.2 \times 10^{- 10}$ atmosphere. The number of molecules in the gas are:

6.02 \times 10^{14}

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8.2 \times 6 \times 10^{24}

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2.46 \times 10^{10}

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6.02 \times 10^{23}

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Solution

The correct option is A $6.02 \times 10^{14}$
$n = \frac{P V}{R T}$
Number of molecules $= n \times N_{o}$
$n = \frac{8.2 \times 10^{- 10} \times 30}{0.082 \times 300} = 10^{- 9}$
Number of molecules $= 10^{- 9} \times 6.02 \times 10^{23} = 6.02 \times 10^{14}$

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A gas is present in a 30 litre vessel at 27oC and a pressure of 8.2×10−10 atmosphere. The number of molecules in the gas are:

The correct option is A 6.02×1014n=PVRTNumber of molecules =n×Non=8.2×10−10×300.082×300=10−9Number of molecules =10−9×6.02×1023=6.02×1014

A gas is present in a 30 litre vessel at $27^{o} C$ and a pressure of $8.2 \times 10^{- 10}$ atmosphere. The number of molecules in the gas are:

The correct option is A $6.02 \times 10^{14}$
$n = \frac{P V}{R T}$
Number of molecules $= n \times N_{o}$
$n = \frac{8.2 \times 10^{- 10} \times 30}{0.082 \times 300} = 10^{- 9}$
Number of molecules $= 10^{- 9} \times 6.02 \times 10^{23} = 6.02 \times 10^{14}$