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Vapour density of ammonia is 8.5 , 85 grams of NH3 at NTP occupy

Sudharsan , 8 Years ago
Grade 11
anser 2 Answers
Lakshay Kumar

Last Activity: 8 Years ago

Vapor density *2=molecular massHere, 8.5*2=molecular mass of NH3Molecular mass of NH3=17g/molNow, no. of moles=given mass/molecular massTherefore, no. of moles=85/17=5molesSince one mole of any gas is equal to 22.7 Litre at NTPTherefore, 5 moles of gas will cover a volume =5*22.7=113.5 Litre
Saumya baloni

Last Activity: 7 Years ago

2×v.d = m.w of NH3
2×8.5=m.w of nh3
17.0g=m.w of nh3
Now
No of moles=85/17 =5 moles
So if 1 mole has 22.4l vol
Then 5 moles has 22.4×5 =112.0litre
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