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The volume of a gas at STP is 1.12 x 10 -7 ml . Calculate the number of molecules present in the gas .

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4 years ago

```							At STP, a volume of 22400 mL is occupid by 6.023*1023 molecules of any gas.So, 1.12 * 10-7mL is occupied by 6.023*1023 *1.12 * 10-7*22400-1 moleules or 3.0115 * 1012 molcules.
```
4 years ago
```							At S.T.P 1mole=6.023×10^23 molecules which occupy a volume of 22400ml.Let the number of molecules present in 1.12×10^-7ml be `X`.So, 6.023×10^23 ×1.12*10^-7= X × 22400                            X=6.023×10^23 ×1.12*10^-7 ÷ 22400 Number of molecules (X)=3.01 ×10^12 molecules
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2 years ago
```							Here cc mean ml , so here at stp 22400 ml volume is occupied by 6.023 * 10²³ molecules	To calculate no of molecules in 1.12*10¯7  volume of gas	1.12 * 10¯7​ * 6.023 * 10²³ divided by 22400	No of molecumol is equal to 3.013* 10¹²
```
2 years ago
```							Dear Student,Please find below the solution to your problem.At S.T.P 1mole=6.023×10^23 molecules which occupy a volume of 22400ml.Let the number of molecules present in 1.12×10^-7ml be `X`.So, 6.023×10^23 ×1.12*10^-7= X × 22400X=6.023×10^23 ×1.12*10^-7 ÷ 22400Number of molecules (X)=3.01 ×10^12 moleculesThanks and Regards
```
6 months ago
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