This uses the concept of freezing point depression. When faced with this issue, we use the following equation:
ΔT = i·Kf·m
which translates in english to:
Change in freezing point = vant hoff factor * molal freezing point depression constant * molality of solution
Because the freezing point depression is a colligative property, it does not depend on the identity of the molecules, just the number of them.
Now, we know that molality will be constant, and Kf will be constant, so our only unknown is "i", or the van't hoff factor.
The van't hoff factor is the number of atoms that dissociate from each individual molecule. The higher the van't hoff factor, the more depressed the freezing point will be.
NaCl will dissociate into Na+ and Cl-, so it has i = 2
CaCl2 will dissociate into Ca2+ and 2 Cl-, so it has i = 3
AlBr3 will dissociate into Al3+ and 3 Br-, so it has i = 4
Therefore, AlBr3 will lower the freezing point of water the most.
Answer:
The equation: (NH₄)₂SO₄ = 2NH4(+) + SO4(-2)
The number of moles = 5 g / 132.14 g/mol = 0.038 mol
The number of molecules = 0.038 X 6.022x10^23 = 2.29x10^23
the number of positive ions present in the ammonium sulphate solution:
2 positive ions for every 1 molecule of (NH₄)₂SO₄
so 2 x 2.29x10^23 = 4.58x10^23
the number of negative ions present in the ammonium sulphate solution
1 negative ion for every 1 molecule of (NH₄)₂SO₄
so 1 x 2.29x10^23 = 2.29x10^23
the total number of ions present in the ammonium sulphate solution
4.58x10^23 + 2.29x10^23 = 6.87x10^23
Answer:
74.0 g/mol
Explanation:
Step 1: Write the generic neutralization reaction
HA + NaOH ⇒ NaA + H₂O
Step 2: Calculate the reacting moles of NaOH
At the equivalence point, 33.83 mL of 0.115 M NaOH react.
0.03383 L × 0.115 mol/L = 3.89 × 10⁻³ mol
Step 3: Calculate the moles of HA that completely react with 3.89 × 10⁻³ moles of NaOH
The molar ratio of HA to NaOH is 1:1. The reacting moles of HA is 1/1 × 3.89 × 10⁻³ mol = 3.89 × 10⁻³ mol.
Step 4: Calculate the molar mass of the acid
3.89 × 10⁻³ moles of HA have a mass of 0.288 g.
M = 0.288 g / 3.89 × 10⁻³ mol = 74.0 g/mol
Explanation:
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