Mass of solute = 25.8 g
mass solution = 212 g
% =( mass of solute / mass solution ) x 100
% = ( 25.8 / 212 ) x 100
% = 0.122 x 100
= 12.2 %
Answer:
16.4 °C
Explanation:
Boiling point elevation is the phenomenon in which the boiling point of a solvent will increase when another compound is added to it; meaning that athe resultant solution has a higher boiling point than its pure solvent.
Using the ebullioscopic constant,
ΔT = m * i * Kb
Where,
Δ T is the temperature difference between the boiling point of the solution, Temp.f and boiling point of the pure solvent, Temp.i
Kb is the ebulliscope factor of water = 0.510 °C.kg/mol
i is the van hoffs number = 1
m is the molality in mol/kg.
Calculating the molality of the solution,
Temp.i = 100°C
Temp.f = 104.5 °C
= 4.5/(1*0.510)
= 8.8235 mol/kg
Freezing point depression is defined as the decrease in the freezing point of a solvent on the addition of a solute.
Using the same equation, but kf = 1.86 °C.kg/mol
ΔT = m * i * Kf
Temp.i = freezing point of water = 0°C
Temp.f = (8.8235*1.86) - 0
= 16.412 °C
Freezing point of the solution = 16.4 °C
SrSo4 = Sr(2+) + SO4(2-)
Let’s say that the initial concentration of SrSo4 was 1. ( or we have 1 mole of this reagent).
When The reaction occurs part of SrSo4is dissociated. And we get X mole Sr(2+) and So4(2-).
Ksp=[Sr(2+)]*[SO4(2-)]
X^2=3.2*10^-7
X=5.6*10^-4
Answer:
N2H4 + 2H2O2 ---->N2 + 4H2O
Explanation:
N=2 N=2
H=6 ->8 H=2 ->8
O=2 -> 4 O=1 -> 4
Add coefficients to hydrogen peroxide on the left and water on the right, so that there is an equal number of hydrogens and oxygens.
