A substance X contains 10 gram of calcium carbonate calculate the number of mole of calcium carbonate present in X
2 answers:
☃️ Chemical formulae ➝
For solving this question, We need to know how to find moles of solution or any substance if a certain weight is given.
Atomic weight of elements:
Ca = 40
C = 12
O = 16
❍ Molecular weight of
= 40 + 12 + 3 × 16
= 52 + 48
= 100 g/mol
❍ Given weight: 10 g
Then, no. of moles,
⇛ No. of moles = 10 g / 100 g mol‐¹
⇛ No. of moles = 0.1 moles
☄ No. of moles of Calcium carbonate in that substance = <u>0.1 moles</u>
<u>━━━━━━━━━━━━━━━━━━━━</u>
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Answer:
3.18 mol
Explanation:
n(CO2) = mass/ Mr.
= 25.5 / 16
= 1.59 mol
As per the equation above,
n(LiOH) : n(CO2)
2 : 1
∴ 3.18 : 1.59
Answer:
We can not swim in 1.00 × 10²⁷ molecules of water
Explanation:
The given number of molecules of water = 1.00 × 10²⁷ molecules
The Avogadro's number, , gives the number of molecules in one mole of a substance
≈ 6.0221409 × 10²³ molecules/mol
Therefore
Therefore, we have;
The number of moles of water present in 1.00 × 10²⁷ molecules, n = (The number of molecules of water) ÷
∴ n = (1.00 × 10²⁷ molecules)/(6.0221409 × 10²³ molecules/mol) = 1,660.53902857 moles
The mass of one mole of water = The molar mass of water = 18.01528 g/mol
The mass, 'm', of water in 1,660.53902857 moles of water is given as follows;
Mass = (The number of moles of the substance) × (The molar mass of the substance)
∴ The mass of the water in the given quantity of water, m = 1,660.53902857 moles × 18.01528 g/mol ≈ 29.9150756 kg.
The density pf water, ρ = 997 kg/m³
Volume = Mass/Density
∴ The volume of the water present in the given quantity of water, v = 29.9150756 kg/(997 kg/m³) ≈ 30.0050909 liters
The volume of the water present in 1.00 × 10²⁷ molecules of water ≈ 30.0 liters
The average volume of a human body = 62 liters
Therefore, we can not swim in the given quantity of 1.00 × 10²⁷ molecules = 30.0 liters water