If 55.6 mL of BaCl2 solution is needed to precipitate all the sulfate ion in a 756 mg sample of Na2SO4, what is the molarity of the solution? Express the molarity to three significant digits.
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To find the molarity of the BaCl2 solution, we need to know the number of moles of BaCl2 used.
First, we need to calculate the number of moles of Na2SO4 in the sample. We have a mass of 756 mg, so we need to convert this to grams:
756 mg = 756/1000 g = 0.756 g
To find the number of moles, we use the formula:
moles = mass / molar mass
The molar mass of Na2SO4 can be calculated by adding up the atomic masses of sodium (Na), sulfur (S), and oxygen (O):
molar mass of Na2SO4 = 2*22.99 g/mol + 32.07 g/mol + 4*16.00 g/mol = 142.04 g/mol
moles of Na2SO4 = 0.756 g / 142.04 g/mol ≈ 0.00532 mol
The balanced chemical equation for the reaction between BaCl2 and Na2SO4 is:
BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq)
From the equation, we can see that 1 mole of BaCl2 reacts with 1 mole of Na2SO4. Therefore, the number of moles of BaCl2 used is also equal to 0.00532 mol.
Molarity is defined as moles of solute per liter of solution. In this case, the number of moles of BaCl2 is 0.00532 mol, and the volume of the solution is 55.6 mL, which is equal to 0.0556 L.
Molarity = moles / volume
Molarity = 0.00532 mol / 0.0556 L ≈ 0.0959 M
Therefore, the molarity of the BaCl2 solution is approximately 0.096 M.
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