A.water.
An aqueous solution is a solution in water.
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Answer:
0.147 mol
Explanation:
Step 1: Calculate the volumetric concentration (Cv)
We will use the following expression.
Cv = Cg × ρ
Cv = 98.0 g%g × 1.84 g/mL = 180 g%mL
Step 2: Calculate the molarity of sulfuric acid
We will use the following expression.
M = mass solute / molar mass solute × liters of solution
M = 180 g / 98.08 g/mol × 0.100 L = 18.4 M
Step 3: Calculate the moles of solute in 8.00 mL of solution
8.00 × 10⁻³ L × 18.4 mol/L = 0.147 mol
Answer:
Mole fraction for solute = 0.1, or 10%
Molality = 6.24 mol/kg
Explanation:
22.3% by mass → In 100 g of solution, we have 22.3 g of HCOOH
Mass of solution = 100 g
Mass of solute = 22.3 g
Mass of solvent = 100 g - 22.3g = 77.7 g
Let's convert the mass to moles
22.3 g . 1mol/ 46 g = 0.485 moles
77.7 g. 1mol / 18 g = 4.32 moles
Total moles = 4.32 moles + 0.485 moles = 4.805 moles
Xm for solute = 0.485 / 4.805 = 0.100 → 10%
Molality → mol/ kg → we convert the mass of solvent to kg
77.7 g. 1 kg / 1000g = 0.0777 kg
0.485 mol / 0.0777 kg = 6.24 m
It is the mass of one mole of the substance.
