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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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oksian1 [2.3K]

Answer:

The standard cell potential is 2.00 V

Explanation:

Step 1: Data given

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Al3++3e−→Al E∘=−1.66 V anode

Cu2++2e−→Cu E∘=0.340 V cathode

Step 2: Balance both equations

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Given - Number of moles of Al(NO3)3 - 4 moles
Number of moles of NaCl - 9 moles
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