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`         A 2.0 g sample of a mixture containing sodium carbonate sodium bicarbonate and sodium sulphate is gently heated till the evolution of CO2 ceases. The volume of CO2 at 750 mm Hg pressure and at 298 K is measured to be 123.9 mL. A 1.5 g of the same sample requires 150 mL of (M/10) HCI for complete neutralization. Calculate the percentage composition of the components of the mixture. `
3 years ago

Deepak Patra
474 Points
```										Sol. CO2 is evolved due to following reaction :
2NaHCO3 → Na2CO3 + H2O + CO2
Moles of CO2 produced= pV/RT
= 750/760 × 123.9/1000 × 1/(0.082 ×298)
= 5 × 10-3
⇒ Moles of NaHCO3 in 2g sample = 2 × 5 × 10-3
⇒ millimol of NaHCO3 in 1.5 g sample
= 0.01/2 × 1.5 × 1000
= 7.5
Let the 1.5 g sample contain x millimol Na2CO3, then 2x + 7.5 = millimol of HCI = 15
⇒ x = 3.75
⇒ Mass of NaHCO3 = (7.5 × 84)/1000 = 0.63 g
Mass of Na2CO3 = (3.75 × 106)/1000 = 0.3975 g
⇒ % mass of NaHCO3 = 0.63/1.50 × 100 = 42 %
% mass of Na2CO3 = 0.3975/1.5 × 100 = 26.5%

```
3 years ago
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