JEE Class main Answered
Sir plz solve it step by step ,U may leave part (b) but plz solve explaining part(a)
![question image](https://images.topperlearning.com/topper/new-ate/topr_291105392745277201908080856341.jpeg)
Asked by vishakhachandan026 | 08 Aug, 2019, 10:21: AM
Given:
For NO,
Volume = 0.25 L
Pressure = 800 torr
![equals space 800 over 760
equals 1.05 space atm](https://images.topperlearning.com/topper/tinymce/cache/aad9a191f64e9ab1f01058e32d9de925.png)
temperature = 220 K
By using gas equation, no. of moles can be calculated as,
PV = nRT
![straight n equals PV over RT
space equals fraction numerator 1.05 cross times 0.25 over denominator 0.0821 cross times 220 end fraction
space straight n equals 0.0145 space mol](https://images.topperlearning.com/topper/tinymce/cache/33d5adc5f36e2ef84ca7bba70018c881.png)
For O2
For NO,
Volume = 0.1 L
Pressure = 600 torr
![equals space 600 over 760
equals 0.79 space atm](https://images.topperlearning.com/topper/tinymce/cache/9b9cb1b360ac681cc2dd588a85b81ff3.png)
temperature = 220 K
By using gas equation, no. of moles can be calculated as,
PV= nRT
![straight n equals PV over RT
space space equals fraction numerator 0.79 cross times 0.1 over denominator 0.0821 cross times 220 end fraction
straight n equals space 0.00437 space mol](https://images.topperlearning.com/topper/tinymce/cache/ab12cddd07d97c7d365a656762f624aa.png)
The reaction to form N2O4
2NO + O2 → N2O4
2 mole of NO reacts with 1 mole of O2
so, 0.00437 mole of O2 will require
![fraction numerator space 0.00437 cross times 2 over denominator 1 end fraction
equals 0.00874 space mol space of space NO](https://images.topperlearning.com/topper/tinymce/cache/d1bdca57d0a671527eb99ddbf7add025.png)
(a)
Excess of NO is : 0.0.145 - 0.00874 = 0.00578 mole of NO
Now aain by using gas law we can calculate pressure,
PV = nRT
![straight P equals nRT over straight V
space space space space equals fraction numerator 0.00576 cross times 0.00821 cross times 220 over denominator open parentheses 0.25 plus 0.1 close parentheses end fraction
straight P equals 0.299 space atm](https://images.topperlearning.com/topper/tinymce/cache/04647b8b42e2889f2cfc69f9ec6b4ebb.png)
(b) From equation,
2 mole of NO gives 1 mole of N2O4
so 0.00874 mole of NO will give,
![equals fraction numerator 0.00874 cross times 1 over denominator 2 end fraction
equals 0.00437 space mole space of space straight N subscript 2 straight O subscript 4](https://images.topperlearning.com/topper/tinymce/cache/94ee048fd42f291134f5ca5362171aef.png)
Molar mass of N2O4 = 92g/mol
Mass of N2O4 = 92×0.00437
= 0.402 gm.
Answered by Varsha | 08 Aug, 2019, 12:52: PM
JEE main - Chemistry
Asked by ruchisharmatbn | 06 Apr, 2024, 08:42: AM
JEE main - Chemistry
Asked by arjuns94037 | 07 Jan, 2024, 12:43: PM
JEE main - Chemistry
Asked by vishakhachandan026 | 08 Aug, 2019, 10:21: AM
JEE main - Chemistry
Asked by ashutosharnold1998 | 08 Aug, 2019, 00:11: AM
JEE main - Chemistry
Asked by vishakhachandan026 | 04 Jul, 2019, 14:59: PM
JEE main - Chemistry
Asked by Anish | 14 Mar, 2019, 11:14: AM
JEE main - Chemistry
Asked by Anish | 09 Jan, 2019, 23:33: PM