In Carius method, 0.75 g of an organic compound gave 1.2 g of barium sulphate, find percentage of sulphur ( m o l a r m a s s 32 g m o l ? ¹ ) . Molar mass of barium sulphate is 233 g m o l ? ¹ . [2026]

Question

In Carius method, 0.75 g of an organic compound gave 1.2 g of barium sulphate, find percentage of sulphur $(molar mass 32 g mol⁻¹).$ Molar mass of barium sulphate is $233 g mol⁻¹ .$ [2026]

Smart Achievers · Accepted Answer

(2)

$\frac{n_{{BaSO}_{4}} \times 32}{W_{(unknown comp .)}} \times 100$

$= \frac{1.2 \times 32}{233} \times \frac{100}{0.75} = 21.97 %$

Solution

Q.
In Carius method, 0.75 g of an organic compound gave 1.2 g of barium sulphate, find percentage of sulphur $(molar mass 32 g mol⁻¹).$ Molar mass of barium sulphate is $233 g mol⁻¹ .$ [2026]

1 4.55%

2 21.97%

3 16.48%

4 10.30%

Ans.
(2)

$\frac{n_{{BaSO}_{4}} \times 32}{W_{(unknown comp .)}} \times 100$

$= \frac{1.2 \times 32}{233} \times \frac{100}{0.75} = 21.97 %$

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