1. The heat of solution of ammonium trioxonitrate (v) solution is 25KJmol. Calculate the quantity of heat absorbed when: (a) 2 moles (b) 5g of ammonium trioxonitrate (v) solid is dissolved with a large volume of water.
Answers
Answer:
For 2 moles , ΔH = 12.5 KJ
For 5 gm , ΔH = 400KJ .
Explanation:
Given :- heat of solution :- 25KJmol
To find :-
a) heat absorbed when 2 moles
b) heat absorbed when 5 gm of ammonium trioxonitrate (v) is dissolved in large amount of water .
Solution :-
a) Formula to be used :- ΔH = heat of solution / no. of moles .
Therefore , n = 2 , q = 25KJmol .
heat absorbed = 25 / 2 = 12.5 KJ
b) Molecular weight of ammonium trioxonitrate(v) = 80 g/mol .
Therefore , no. of moles (n) = 5 / 80 = 0.0625
Heat absorbed = 25/0.0625 = 400 KJ .
#SPJ2
Answer:
ΔH For 2 Moles = 12.5KJ
ΔH For 5 g = 400 KJ
Explanation:
GIVEN :
Heat of solution of ammonium-trioxonitrate (v) = 25KJ/mol
TO FIND :
(a) Heat absorbed when 2 moles of ammonium trioxonitrate is dissolved in water .
(b) Heat absorbed when 5 g of ammonium trioxonitrate is dissolved in water .
Step by Step Explanation :
Heat absorbed (ΔH) = heat of solution / no. of moles
(a) no. of moles = 2
∴ ΔH = (25 / 2) KJ
ΔH = 12.5KJ
(b) no. of moles = given wt. / molar mass of ammonium trioxoxnitrate
= 5 / 80 ( 5 = given wt. and 80 = molar mass )
= 0.0625 mol
∴ ΔH = 25 / 0.0625
ΔH = 400 KJ
#SPJ2