From the calculations, we can see that, the change in the freezing point is -0.634°C.
<h3>What is freezing point?</h3>
The term freezing point refers to the temperature at which a liquid is changed to solid.
Given that;
ΔT = K m i
Number of moles sucrose = 35.0 g/ 342.30 g/mol = 0.1 moles
molality = 0.1 moles/ 300.0 * 10^-3 Kg
= 0.33 m
Thus;
ΔT = -1.86°C/mol * 0.33 m * 1
= -0.634°C
Learn more about freezing point:brainly.com/question/3121416
#SPJ1
Explanation:
The unequal sharing of electrons between the atoms and the unsymmetrical shape of the molecule means that a water molecule has two poles - a positive charge on the hydrogen pole (side) and a negative charge on the oxygen pole (side). We say that the water molecule is electrically polar.
So in your question that ask to calculate the Ph result of the resulting solution if 26 ml of 0.260 M HCI(aq) is added to the following substance. The the result are the following:
A. The result is pH= 14-pOH
B. There are 10ml of 0.26m HCL excees in this reaction so the answer is log(H)+
26.7% is the percent composition by mass of sulfur in a compound named magnesium sulfate. Explanation: Molar mass of compound = 120 g/mol.
Answer is: the freezing point is 1.63°C and boiling point is 82.01°C.<span>.
1) n(</span><span>nonelectrolyte solute) = 0.656 mol.
</span>m(C₆H₆ - benzene) = 869 g ÷ 1000 g/kg.
m(C₆H₆) = 0.869 kg.<span>
b(solution) = n(</span>nonelectrolyte solute) ÷ m(C₆H₆).<span>
b(solution) = 0.656 mol ÷ 0.869 kg.
b(solution) = 0.754 mol/kg.
2) ΔT = Kf(benzene) · b(solution).
ΔT = 5.12°C/m · 0.754 m.
ΔT = 3.865°C.
Tf = 5.50°C - 3.865°C.
Tf = 1.63°C.
</span>
3) ΔTb = Kb(benzene) · b(solution).
ΔTb = 2.53°C/m · 0.754 m.
ΔTb = 1.91°C.
Tb = 80.1°C + 1.91°C.
Tb = 82.01°C.<span>
</span>
