How many moles of Na are present in 120 g Sodium sulphate.

1.	How many moles of Na are present in 120 g Sodium sulphate.
Answer» Molar mass of Sodium sulphate (Na₂SO₄)- = 2(23)+32+4(16) = 46+32+64 = 142 g/mol Number of moles of Na₂SO₄ in 120 grams of it: No. of moles = Molecular mass/ Molar mass No. of moles = 120/142 = 0.845 moles Now, in each mole of sodium sulphate, there are 2 moles of Sodium (Na). Therefore, 0.845×2 = 1.69 moles of sodium are present in 120 grams of sodium sulphate. Na₂SO₄(sodium sulphate) Molecular weight - 142 g/mol Number of moles of sodium sulphate = \(\frac{120g}{142g/mol} = 0.84 mole\) \(\because\) 1 molecule of Na₂SO₄ contains = 2 atom of sodium \(\therefore\) 0.84 mole of Na₂SO₄ contain = 1.68 moles sodium atom. Hence, 120 g Na₂SO₄ contain 1.68 moles sodium atom

Answer»

Molar mass of Sodium sulphate (Na₂SO₄)-

= 2(23)+32+4(16)

= 46+32+64

= 142 g/mol

Number of moles of Na₂SO₄ in 120 grams of it:

No. of moles = Molecular mass/ Molar mass

No. of moles = 120/142 = 0.845 moles

Now, in each mole of sodium sulphate, there are 2 moles of Sodium (Na). Therefore, 0.845×2 = 1.69 moles of sodium are present in 120 grams of sodium sulphate.

Na₂SO₄(sodium sulphate)

Molecular weight - 142 g/mol

Number of moles of sodium sulphate = \(\frac{120g}{142g/mol} = 0.84 mole\)

\(\because\) 1 molecule of Na₂SO₄ contains = 2 atom of sodium

\(\therefore\) 0.84 mole of Na₂SO₄ contain = 1.68 moles sodium atom.

Hence, 120 g Na₂SO₄ contain 1.68 moles sodium atom

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