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1 year ago

What is the molarity of a solution that contains 2.35 g of nh3 in 0.0500 l of solution?

Chemistry

2 answers:

torisob [31]1 year ago

5 0

Answer:

M = 2.76 M

Explanation:

To calculate the molarity of any solution, we need to use the following expression:

M = n/V (1)

Where:

M: Molarity of solution (mol/L or M)

n: moles of NH3 (moles)

V: volume of solution (In Liters)

But we do not have the moles of NH3, so it needs to be calculated. We have the mass and we can use the following expression to calculate the moles:

n = m/MM (2)

Where:

m: mass of NH3 (g)

MM: molar mass of NH3 (g/mol)

The reported molar mass of NH3 is 17.031 g/mol, so, using the mass of NH3 and this MM, we can calculate the moles:

n = 2.35 / 17.031 = 0.138 moles

With these moles, we can calculate the molarity:

M = 0.138 / 0.05

M = 2.76 mol/L

And this is the concentration or molarity of the NH3 solution.

tankabanditka [31]1 year ago

3 0

The molarity is the number of moles in 1 L of the solution.
The mass of NH₃ given - 2.35 g
Molar mass of NH₃ - 17 g/mol
The number of NH₃ moles in 2.35 g - 2.35 g / 17 g/mol = 0.138 mol
The number of moles in 0.05 L solution - 0.138 mol
Therefore number of moles in 1 L - 0.138 mol / 0.05 L x 1L = 2.76 mol
Therefore molarity of NH₃ - 2.76 M

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Here comes the concept of Limiting reagent:

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