Question

112 ml of an NO2 at 1 ATM and 273 K was liquefied, the density of liquid being 1.15 gram per ml calculate the volume of and the number of molecules in the liquid and NO2

Answer 1

Answer:

112 ml of an NO2 at 1 ATM and 273 K was liquefied

We know, At STP 22400ml of NO2=46g of NO2

According to the problem:

112.0ml of NO2=112.0mL×46.0g22400mL=0.23g

As we know, Volume of NO2=Mass/Density

   Therefore                           = 0.23g/1.15g ml^-1 = 0.2ml

Hence the Volume is 0.2ml

Number of molecules = 0.23/46* 6.023* 10^23

⇒3.01×10^21

Hence the number of molecules are 3.01×10^21

Answer 2

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