In order to determine the concentration of ammonium ions in
the solution prepared by mixing solutions of ammonium sulfate, (NH4)2SO4, and ammonium
nitrate, first calculate the amount of ammonium ions for each solution.<span>
<span>For ammonium sulfate sol'n: 0.360 L x 0.250 mol(NH4)2SO4/ L x 2 mol NH4+ /1 mol(NH4)2SO4 =
0.18 mol NH4+
<span>For ammonium nitrate sol'n: 0.675 x 1.2 mol NH4NO3/L x 1 mol NH4+ /1 molNH4NO3
= 0.81 mol NH4+
Thus, the amount of NH4+ ions is (0.18 + 0.81) mol or 0.99
mol NH4+. To get the concentration, multiply this to the volume of solution
which is assumed to be additive, such that:</span></span></span>
M NH4+ in sol’n = 0.99 mol NH4+/1.035 L = 0.9565 mol NH4+/ L
sol’n
Given is the specific heat of water equal to 4.18 Joule per gram per *C.
This means to raise the temperature of 1 g of water by 1 degree Celsius we need 4.18 joule of energy.
Now, look at the question. We are asked that how much amount of energy would be required to raise the temperature of 25 g of water by (54-50) = 4 degree celsius.
To do so we have formula
Q = m C (temperature difference)
Have a look at pic for answer
Ok to answer this question we firsst need to fin the number of mol of Urea (CH4N2O). to do this we simply :
1 mol of urea =15/60.055 = 0.25mol
therefore 200g of water contain 0.25mol
the next step is to determine the malality of our solution in 200g of water, to do this we say:
200 g = 1Kg/1000g = 0.2kg
therefor 0.25mol/0.2Kg = 1.25mol/kg
and from the equation:
we know that i = 1
we are given Kf
b is the molality that we just calculated
therefore;
the solutions freezing point is -2.325°C