3.5 M has 3.5 moles per litre
so we have one litre, so we need 3.5 moles
moles = mass/molarmass
3.5 * 23 = 80.5
Answer:0.477 g/ml
Explanation:
Density=(40.14-33.79)/13.3 ml
Density=6.35/13.3
Density=0.477 g/ml
Answer:
0.245 moles
Explanation:
moles of lithium = 2*moles of Li2SO4
so moles of LI2SO4=2.94*10^23/(2*6*10^23)=0.245
Answer: 167 g
Explanation:
1) The depression of the freezing point of a solution is a colligative property ruled by this equation:
ΔTf = i × m × Kf
Where:
ΔTf is the decrease of the freezing point of the solvent due to the presence of the solute.
i is the Van't Hoof factor and is equal to the number of ions per each mole of solute. It is only valid for ionic compounds. Here the solute is not ionice, so you take i = 1
Kf is the molal freezing constant and is different for each solvent. For water it is 1.86 m/°C
2) Calculate the molality (m) of the solution
ΔTf = i × m × Kf ⇒ m = ΔTf / ( i × Kf) = 5.00°C / 1.86°C/m = 2.69 m
3) Calculate the number of moles from the molality definition
m = moles of solute / kg of solvent ⇒ moles of solute = m × kg of solvent
moles of solute = 2.69 m × 1.00 kg = 2.69 moles
4) Convert moles to grams using the molar mass
molar mass of C₂H₆O₂ = 62.07 g/mol
mass in grams = number of moles × molar mass = 2.69 moles × 62.07 g/mol = 166.97 g ≈ 167 g
Answer:
Hydrosulfuric acid will act as limiting reactant.
Explanation:
Given data:
Mass of iron(III) chloride = 3243.0 g
Mass of hydrosulfuric acid = 511.8 g
Limiting reactant = ?
Solution:
Chemical equation:
2FeCl₃ + 3H₂S → Fe₂S₃ + 6HCl
Number of moles of iron(III) chloride:
Number of moles = mass/molar mass
Number of moles = 3243.0 g/ 162.2 g/mol
Number of moles = 20 mol
Number of moles of hydrosulfuric acid:
Number of moles = mass/molar mass
Number of moles = 511.8 g/ 34.1 g/mol
Number of moles = 15 mol
Now we will compare the moles of both reactant with products
FeCl₃ : Fe₂S₃
2 : 1
20 : 1/2 ×20 = 10
FeCl₃ : HCl
2 : 6
20 : 6/2 ×20 = 60
H₂S : Fe₂S₃
3 : 1
15 : 1/3 ×15 = 5
H₂S : HCl
3 : 6
15 : 6/3 ×15 = 30
Hydrosulfuric acid producing less number of moles of product thus, it will act as limiting reactant.
