Answer:
15.75 grams of HNO3 was used and dissolved in 2.5 liters of solvent, to make a 0.10 M solution
Explanation:
Step 1: Data given
Nitric acid = HNO3
Molar mass of H = 1.01 g/mol
Molar mass of N = 14.0 g/mol
Molar mass O = 16.0 g/mol
Number of moles nitric acid (HNO3) = 0.25 moles
Molairty = 0.10 M
Step 2: Calculate molar mass of nitric acid
Molar mass HNO3 = Molar mass H + molar mass N + molar mass (3*O)
Molar mass HNO3 = 1.01 + 14.0 + 3*16.0
Molar mass HNO3 = 63.01 g/mol
Step 3: Calculate mass of solute use
Mass HNO3 = moles HNO3 * molar mass HNO3
Mass HNO3 = 0.25 moles * 63.01 g/mol
Mass HNO3 = 15.75 grams
15.75 grams of HNO3 was used and dissolved in 2.5 liters of solvent, to make a 0.10 M solution
The molecular weight of unknown gas : 23.46 g/mol
<h3>Further explanation</h3>
Given
A vessel contains 10% of oxygen and 90% of an unknown gas.
diffuses rate of mixed gas = 86 s
diffuses rate of O₂ = 75 s
Required
the molecular weight of unknown gas (M)
Solution
The molecular weight of mixed gas :(M O₂=32 g/mol)
Graham's Law :
Answer:
a) Molar fraction:
Methane: 67,5%
Ethane: 25,1%
Propane: 7,4%
b) Molecular weight of the mixture: 25,16 g/mol
Explanation:
with a basis of 100 kg:
Moles of methane:
500 g ×
= 31,2 moles
Mass of ethane
350 g ×
= 11,6 moles
Mass of propane
150 g ×
= 3,4 moles
Total moles: 31,2 moles + 11,6 moles + 3,4 moles = <em>46,2 moles</em>
Molar fraction of n-methane:
=67,5%
Molar fraction of ethane:
= 25,1%
Molar fraction of propane:
= 7,4%
b) Average molecular weight:
= <em>25,16g/mol</em>
I hope it helps!
the molecular interpretation is not suitable for the assumption that there are much more interactions in the intermolecular level
We must account for the breaking/creation of hydrogen bondings, which is not the scope of the equilibria made in trouton's analysis
