Answer:
334.2× 10²³ molecules
Explanation:
Given data:
Mass of water = 1 Kg ( 1000 g )
Number of molecules = ?
Solution:
Number of moles of water:
Number of moles = mass/ molar mass
Number of moles = 1000 g/ 18 g/mol
Number of moles = 55.5 mol
1 mole contain 6.022× 10²³ molecules
55.5 mol×6.022× 10²³ molecules
334.2× 10²³ molecules
<u>Ans: Acetic acid = 90.3 mM and Sodium acetate = 160 mM</u>
Given:
Acetic Acid/Sodium Acetate buffer of pH = 5.0
Let HA = acetic acid
A- = sodium acetate
Total concentration [HA] + [A-] = 250 mM ------(1)
pKa(acetic acid) = 4.75
Based on Henderson-Hasselbalch equation
pH = pKa + log[A-]/[HA]
[A-]/[HA] = 10^(pH-pKa) = 10^(5-4.75) = 10^0.25 = 1.77
[A-] = 1.77[HA] -----(2)
From (1) and (2)
[HA] + 1.77[HA] = 250 mM
[HA] = 250/2.77 = 90.25 mM
[A-] = 1.77(90.25) = 159.74 mM
Answer:
The answer to your question: 0.7 M
Explanation:
Data
V of KOH = 90 ml
[KOH] = ?
V H2SO4 = 21.2 ml
[H2SO4] = 1.5 M
2KOH(aq) + H₂SO₄(aq) → K₂SO₄(aq) + 2H₂O(l)
Molarity = moles / volume
moles of H₂SO₄ = (1.5) (21.2)
= 31.8
2 moles of KOH -------------- 1 mol of H₂SO₄
x -------------- 31.8 mol of H₂SO₄
x = (31.8)(2) / 1
x = 63.8 moles of KOH
Molarity = 63.8 / 90
= 0.7 M
Explanation:
Moles of NaOH = 10g / (40g/mol) = 0.25mol.
0.25mol / 500g = 0.50mol / 1000g = 0.50mol/dm³.
The molarity is 0.50mol/dm³.
