Answer:
2.7 °C.kg/mol
Explanation:
Step 1: Calculate the freezing point depression (ΔT)
The normal freezing point of a certain liquid X is-7.30°C and the solution freezes at -9.9°C instead. The freezing point depression is:
ΔT = -7.30 °C - (-9.9 °C) = 2.6 °C
Step 2: Calculate the molality of the solution (b)
We will use the following expression.
b = mass of solute / molar mass of solute × kilograms of solvent
b = 102. g / (162.2 g/mol) × 0.650 kg = 0.967 mol/kg
Step 3: Calculate the molal freezing point depression constant Kf of X
Freezing point depression is a colligative property. It can be calculated using the following expression.
ΔT = Kf × b
Kf = ΔT / b
Kf = 2.6 °C / (0.967 mol/kg) = 2.7 °C.kg/mol
Answer:
The molar mass is: 18.02 g/mol.
Explanation:
- Mass of two moles of Hydrogen atoms (H2) = 2x 1 g/mol = 2 g/mol.
- Mass of one mole of water (H2O) = 2 g/mol + 16 g/mol = 18 g/mol.
1 mole of Hydrogen= 1.01, so if we have 2 moles of it here, that would be 2.02.
1 mole of Oxygen (that's all we have here)= 16.00
Once you add the two together (2.02+16.00), you will get 18.02.
I hope this made sense! Have a great day!
Answer:
HPO42- (aq) + NH4+ (aq)
Explanation:
This is a conjugate acid-base pair. Please forgive if my answers are incorrect. I myself am quite unsure.
