Answer:
19.8 %
Explanation:
V% = ( V of solute \ 100 ml of solution ) ×100% =
the volume of the solute in 100ml solition = 20 ml
42.9 ml of solute ----------215 ml solution
× ----------- 100 ml solution
42.6 × 100 ml ÷ 215 ml
= 19.8 ml of solute
the V% = (19.8 ÷ 100) ×100% = 19.8%
Answer : The amount of heat released, 45.89 KJ
Solution :
Process involved in the calculation of heat released :
Now we have to calculate the amount of heat released.
![Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+\Delta H_{fusion}+[m\times c_{p,s}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=Q%3D%5Bm%5Ctimes%20c_%7Bp%2Cl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2B%5CDelta%20H_%7Bfusion%7D%2B%5Bm%5Ctimes%20c_%7Bp%2Cs%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
Q = amount of heat released = ?
m = mass of water = 27 g
= specific heat of liquid water = 4.184 J/gk
= specific heat of solid water = 2.093 J/gk
= enthalpy change for fusion = 40.7 KJ/mole = 40700 J/mole
conversion : 
Now put all the given values in the above expression, we get
![Q=[27g\times 4.184J/gK\times (314-273)k]+40700J+[27g\times 2.093J/gK\times (273-263)k]](https://tex.z-dn.net/?f=Q%3D%5B27g%5Ctimes%204.184J%2FgK%5Ctimes%20%28314-273%29k%5D%2B40700J%2B%5B27g%5Ctimes%202.093J%2FgK%5Ctimes%20%28273-263%29k%5D)
(1 KJ = 1000 J)
Therefore, the amount of heat released, 45.89 KJ
Enthalpy change during the dissolution process = m c ΔT,
here, m = total mass = 475 + 125 = 600 g
c = <span>specific heat of water = 4.18 J/g °C
</span>ΔT = 7.8 - 24 = -16.2 oc (negative sign indicates that temp. has decreases)
<span>
Therefore, </span>Enthalpy change during the dissolution = 600 x 4.18 X (-16.2)
= -40630 kJ
(Negative sign indicates that process is endothermic in nature i.e. heat is taken by the system)
Thus, <span>enthalpy of dissolving of the ammonium nitrate is -40630 J/g</span>
