[gecko85]

The heat of solution of KI(s) in water is +20.3 kJ mol–1. A quantity of KI
is added to sufficient water at 23.5 °C in a styrofoam cup to produce 150.0
mL of 2.50 M KI. Calculate the final temperature of the solution.
(Assume a density of 1.30 g mL–1 and a specific heat of 2.7 J g–1 °C–1 for
2.50 M KI).

[MrTeo]

Any ideas?

[gecko85]

delta H=20.3 kJ/mol
density=1.30 g/mL
c=2.7 J/g.C
v=150 mL
M=2.5 mol/L
...T1=23.5 C

density=mass / volume

mass KI=1.30 g/mL x 150 mL
=195 g

enthalpy change

delta H = Q/n

n=MV/1000
=(2.5 mol/L x 150 mL)/1000
=0.375 mol

Q=0.375 mol x 20.3 kJ/mol
=7.61 kJ

delta T=Tf - T1(temperature different cannot be -ve)
since this is endothermic reaction the initial temperature will be greater than temp final.

Q=cm( T1 - Tf )
-Tf= ((7.61 kJ x 1000) / ( 2.7 J/g.C x 195 g)) - 23.5 C

somebody correct me pls....

[MrTeo]

Everything seems OK  

[gecko85]

thank you mr teo... i just answered one question.. but i need somebody to correct me.. sory for my bad english(im trying so hard with it)

Pure oxygen gas was first prepared by heating mercury(II) oxide, HgO:
2 HgO(s) → 2 Hg(l) + O2(g)
What volume (in liters) of oxygen at STP is released by heating 10.57g of
HgO?

answer

mol HgO react to produce 1 mol O2

Mol of HgO = 10.57 g / 216.59 g/mol

= 0.0488
...mol

If 2
mol HgO react to produce 1 mol
O2

0.0488 mol HgO will produce 0.0244 mol O2

At STP
1 mol of gas will occupied 22.4 L

If 0.0244 mol O2 will occupied 0.5466
L

Volume O2 produce = 0.5466 L

[Borek]

0.5466 L

Looks OK.

[gecko85]

thank you very much borek