Molarity: moles of solute dissolved/liter of solution Mole = 6.02 times 10^23 at

Question

Molarity: moles of solute dissolved/liter of solution Mole = 6.02 times 10^23 atoms = Avogadro's number Mole = gram formula weight (FW) or gram molecular weight (MW) These values are usually found on the labels or in Periodic Table. e.g. Na2SO4 2 sodium atoms - 2 times 22.99g = 45.98 g 1 sulfur atom - 1 times 32.06g = 32.06 g 4 oxygen atoms - 4 times 16 00g = 64.00 Total = 142.04g 1M = 1 mole/liter, 1 mM = 1 millimole/liter, 1 uM = umole/liter How much sodium is needed to make 1L of 1M solution? Formula weight of sodium sulfate = 142.04g Dissolve 142.04g of sodium sulfate in about 750mL of water, dissolve sodium sulfate thoroughly and make up the volume to 1 liter (L)

Explanation / Answer

Given:

1 L of 1 M of solution of sodium

Molarity = 1 M = 1 mol/L

Volume = 1 L

No. of moles of sodium = Molarity * Volume = 1 mol/L * 1 L = 1 Mole of sodium

mass of 1 Mole of sodium = 1 mole * Molar mass of sodium = 1mole * 22.99 g/mol = 22.99 g

Given

Mass of sodium sulfate = 142.04 g

Molar mass of sodium sulfate = 142.04 g/mol

No. of moles = Mass / Molar mass = 142.04 g / 142.04 g/mol = 1 mole of sodium sulfate

Volume = 750 mL = 0.750 L ( given)

Molarity = No. of moles sodium sulfate / Volume in solution in Litres = 1 mole / 0.750 L = 1.333 mol/L (or M)

when we make it upto 1L

Molarity = No. of moles of sodium sulfate / volume of solution in Litres = 1 mole / 1 L = 1 mol/L ( or M)

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