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3 years ago

What is the molarity of a solution that contains 87.75g of NaCI in 500.ml of solution

Chemistry

1 answer:

Brut [27]3 years ago

6 0

Answer:

M = 3.0 mol/L.

Explanation:

We can calculate the molarity of a solution using the relation:

M = (mass x 1000) / (molar mass x V)

M is the molarity "number of moles of solute per 1.0 L of the solution.

mass is the mass of the solute (g) (m = 87.75 g of NaCl).

molar mass of NaCl = 58.44 g/mol.

V is the volume of the solution (ml) (V = 500.0 ml).

∴ M = (mass x 1000) / (molar mass x V) = (87.75 g x 1000) / (58.44 g/mol x 500.0 ml) = 3.0 mol/L.

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zlopas [31]

Answer:

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Explanation:

Hello there!

In this case, since this acid-base neutralization is performed in a 1:2 mole ratio of acid to base as the former is a diprotic acid (two hydrogen ions in the molecule), we can write the following equation:

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$M_{acid}=\frac{M_{base}V_{base}}{2V_{acid}}$

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