What is the mass % of Chromium in sample if this sample of chromium containing alloy weighing 3.45g was dissolved in acid and all chromium was oxidised to textCrO42 - . 3.15g of Na2SO3 was required to reduce the textCrO42 - text to textCrO2- in a basic solution while textSO32 - ion being oxidised to textSO42 - . (Molarmass of Cr =52)

Question

What is the mass % of Chromium in sample if this sample of chromium containing alloy weighing 3.45g was dissolved in acid and all chromium was oxidised to textCrO42 - . 3.15g of Na2SO3 was required to reduce the textCrO42 - text to textCrO2- in a basic solution while textSO32 - ion being oxidised to textSO42 - . (Molarmass of Cr =52) (A) 25.13% (B) 76.3% (C) 80% (D) 10%

Tardigrade Admin · Accepted Answer

Gram equivalent of Na2SO3 used up = (3 text. 1 5/6 3) = 5 0 × 1 0- 3 Solution

Moles of textCrO42 - = (5 0 × 1 0- 3/3) = 1 6 text. 6 7 × 1 0- 3 moles of Cr Solution

â Mass of Cr = 16.67 x 10-3 x 52 = 0.867 g % Cr is sample = (0 text. 8 6 7/3 . 4 5) × 1 0 0 = 2 5 text. 1 3 text %

25.13%

76.3%

80%

10%

Gram equivalent of Na2SO3 used up =633.15​=50×10−3 Moles of CrO42−​=350×10−3​=16.67×10−3 moles of Cr ​ Mass of Cr = 16.67 x 10-3 x 52 = 0.867 g % Cr is sample =3.450.867​×100=25.13%

Gram equivalent of Na₂SO₃ used up $= \frac{3 . 15}{63} = 50 \times 1 0^{- 3}$

Moles of $CrO_{4}^{2 -} = \frac{50 \times 1 0 ^{- 3}}{3} = 16.67 \times 1 0^{- 3}$ moles of Cr

Mass of Cr = 16.67 x 10^-3 x 52 = 0.867 g
% Cr is sample $= \frac{0 . 867}{3.45} \times 100 = 25.13 %$