Answer:
[C₆H₁₂O₆] = 0.139 M
Explanation:
Molarity si defined as a sort of concentration. It indicates the moles of solute that are contained in 1 L of solution.
We can also say, that molarity are the mmoles of solute contained in 1 mL of solution.
For this case, the solute is sugar (glucose). Let's determine M (mmol/mL)
(3.95 g . 1mol / 180g) . (1000 mmol / 1mol) / 158 mL
We determine moles, we convert them to mmoles, we divide by mL
M = 0.139 M
Moles = 3.95 g . 1mol / 180g → 0.0219 mol
We convert mL to L → 158 mL . 1L/1000mL = 0.158L
M = 0.0219 mol / 0.158L = 0.139 M
Gasoline would be a mixture if I recalled
Answer:
2.3 * 10^-5
Explanation:
Recall that the solubility of a solute is the amount of solute that dissolves in 1 dm^3 or 1000cm^3 of solution.
Hence;
Amount of calcium oxalate = 154 * 10^-3/128.097 g/mol = 1.2 * 10^-3 mols
From the question;
1.2 * 10^-3 mols dissolves in 250 mL
x moles dissolves in 1000mL
x = 1.2 * 10^-3 mols * 1000/250
x= 4.8 * 10^-3 moldm^-3
CaC2O4(s) ------->Ca^2+(aq) + C2O4^2-(aq)
Hence Ksp = [Ca^2+] [C2O4^2-]
Where;
[Ca^2+] = [C2O4^2-] = 4.8 * 10^-3 moldm^-3
Ksp = (4.8 * 10^-3)^2
Ksp = 2.3 * 10^-5