Question

1.5g of an impure sample of sodium sulphate dissolved in water treated with access of barium chloride solution when 1.74gm of barium sulphate obtained as white ppt form calculate the purity of sample

Answer 1

1.5 g of an impure sample of sodium sulphate is treated with access amount of barium chloride and 1.74gm of barium sulphate obtained as white precipitate form.

chemical reaction is....

$Na_2SO_4+BaCl_2\rightarrow2NaCl+BaSO_4(\downarrow)$

here it is clearly shown that 1 mole of sodium sulphate produces 1 mole of barium sulphate.

molecular weight of Na2SO4 = 23 × 2 + 32 + 64 = 46 + 32 + 64 = 142 g/mol

molecular weight of BaSO4 = 137 + 32 + 64 = 233 g/mol

so, 233 g of BaSO4 is produced by 142 gm of NaSO4

or, 1.74gm of BaSO4 is produced by 142/233 × 1.74 = 1.06gm of Na2SO4

but given mass of Na2SO4 = 1.5gm

so, impurity in sample = 1.5 - 1.06

= 0.42gm

so, purity of sample = 1.06/1.5 × 100 = 70.67 %

Answer 2

Answer:

% age of purity = 70.67 %

Explanation:

equation

Na2SO4 +  BaCl2 → 2NaCl + BaSO4

here it's clear that

1 mol of Na2SO4 = 1 mol of BaSO4

i.e

142g of Na2SO4 produces 233 g of BaSO4

so , x g of Na2SO4 produces 1.74 g BaSO4

But given mass of Na2SO4 = 1.5 g

x = 1.74 X 142/233 = 1.06g

∴ impurity present = 1.5 - 1.06

                               = 0.44 g

∴ Purity % age = 1.06/1.5 X 100 = 70.67 %

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