Answer:
52.2 g
Explanation:
Step 1: Write the balanced equation
3 KOH + H₃PO₄ ⟶ K₃PO₄ + 3 H₂O
Step 2: Calculate the moles corresponding to 89.7 g of KOH
The molar mass of KOH is 56.11 g/mol.
89.7 g × 1 mol/56.11 g = 1.60 mol
Step 3: Calculate the moles of H₃PO₄ needed to react with 1.60 moles of KOH
The molar ratio of KOH to H₃PO₄ is 3:1. The moles of H₃PO₄ needed are 1/3 × 1.60 mol = 0.533 mol.
Step 4: Calculate the mass corresponding to 0.533 moles of H₃PO₄
The molar mass of H₃PO₄ is 97.99 g/mol.
0.533 mol × 97.99 g/mol = 52.2 g
Answer:
Yes
Explanation:
Yes, but almost insignificantly. The regular boiling point of water is 100 degrees celcius, adding a small handfull of salt will increase the boiling point to roughly 100.04 degrees celcius. Therefore, it is almost insignificant but still does affect it. This is mainly because salt changes the way the molecules react thus making it harder for them to change state from liquid to gas. This ultimately causes the water to get hotter before boiling actually occurs.
Answer:
Molar mass for the unknown solute is 109 g/mol
Explanation:
Freezing point depression is the colligative property that must be applied to solve the question.
T°F pure solvent - T°F solution = Kf . m
Let's analyse the data given
Camphor → solvent
Unknown solute → The mass we used is 0.186g
T°F pure solvent = 179.8°C and T°F solution = 176.7°C.
These data help us to determine the ΔT → 179.8°C - 176.7°C = 3.1°C
So we can replace → 3.1°C = 40°C/m . m
m = 3.1°C / 40 m/°C → 0.0775 mol/kg
We have these moles of solute in 1kg of solvent, but our mass of camphor is 22.01 g (0.02201 kg).
We can determine the moles of solute → molality . kg
0.0775 mol/kg . 0.02201 kg = 1.70×10⁻³ moles
Molar mass → mass (g) / moles → 0.186 g / 1.70×10⁻³ mol = 109 g/mol