What is the molarity of a solution that contains 2.35 g of NH3 in 0.0500L of solution?
What is the molarity of a solution that contains 2.35 grams of ammonia (NH3) in 0.0500 liters of solution?
A) 1.00 M
B) 47.0 M
C) 27.6 M
D) 2.76 M
E) 0.0276 M
5 Answers
Feb 20, 2025
You have 2.35g in 0.0500 litres or 1 /0.050*2.35 = 47g per litre
molar mass of NH3 = 17g/mol
147g = 47/17 = 2.765moles per litre
Molarity = 2.76M , answer D
Here is a descriptive solution to this problem:
Given:
Weight of NH3 = 2.35 g
Volume of solution = 0.0500 L
Solution:
Molarity = Number of moles per liter of solution
Molarity = Number of moles / 1L
Number of moles = Weight / Molecular Weight (Mass)
Molecular Weight (Mass) of NH3 = (1)(14) + (3)(1) = 17 g / mole
Number of moles = 2.35 g / 17 g / mole
Number of moles = 0.1382 mole
Molarity = 0.1382 mole / 0.0500 L
Therefore:
Molarity = 2.765 Molar (Prepared in a 50 mL volumetric flask using 2.35 g NH3)
Answer is D).
Good Luck !!!!
You need to find out how many g there are in 1L:
2.35 in 0.05L = 2.35 / 0.05 in 1L = 47g.
The molar mass of ammonia is: 17 g per mole.
47 / 17 = 2.76M.
Answer is D....2.76M
Molarity = (grams of NH3 / M.W. of NH3) / liters of solution
Feb 19, 2025
1 mole NH3 = 1(N) + 3(H) = (1x14) + (1x3) =17g
Number of moles =
mass present in solution/mass of 1mole
=2.35g/17g
=0.138moles
concentration = number of moles/volume
=0.138/0.5 = 0.276/mol/l
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