A 200 g sample of a mixture containing sodium carbonate, sodium bicarbonate and sodium sulphate is gently heated till the evolution of CO 2 ceases. The volume of CO 2 at 2.5 atm pressure and at 30

Question

A 200 g sample of a mixture containing sodium carbonate, sodium bicarbonate and sodium sulphate is gently heated till the evolution of CO 2 ​ ceases. The volume of CO 2 ​ at 2.5 atm pressure and at 300 K is measured to be 3 L . A 200 g of the same sample requires 1.5 L of ( 1 M ) HCl for complete neutralisation. Calculate the mass % of Na 2 ​ CO 3 ​ in mixture. [Atomic mass : Na = 23, S = 32] (Take R = 1/12 L-atm mol -1 K -1 )

TestHub · Accepted Answer

Question Explanation: Find the mass % of Na₂CO₃ in the mixture. Concept: Double titration Solution: On heating, only NaHCO₃ decomposes:

$2 N a H C O_{3} \to N a_{2} C O_{3} + H_{2} O + C O_{2}$

Moles of CO₂ evolved ( $n$ ) = $\frac{P V}{RT} = \frac{2.5 \times 3}{( 1/12 ) \times 300} = \frac{7.5}{25} = 0.3 m o l$

From stoichiometry, moles of NaHCO₃ = $2 \times n (C O_{2}) = 2 \times 0.3 = 0.6 m o l$

For complete neutralization, total moles of HCl = $1.5 L \times 1 M = 1.5 m o l$

Neutralization equations:

$N a_{2} C O_{3} + 2 H Cl \to 2 N a Cl + H_{2} O + C O_{2}$

$N a H C O_{3} + H Cl \to N a Cl + H_{2} O + C O_{2}$

$2 \times m o l eso f N a_{2} C O_{3} + 1 \times m o l eso f N a H C O_{3} = m o l eso f H Cl$

$2 \times n (N a_{2} C O_{3}) + 0.6 = 1.5$

$2 \times n (N a_{2} C O_{3}) = 1.5 - 0.6 = 0.9$

$n (N a_{2} C O_{3}) = \frac{0.9}{2} = 0.45 m o l$

Mass of Na₂CO₃ = moles $\times$ molar mass = $0.45 \times [2 (23) + 12 + 3 (16)] = 0.45 \times 106 = 47.7 g$

Mass % of Na₂CO₃ = $\frac{ma sso f N a _{2} C O _{3}}{t o t a l s am pl e ma ss} \times 100 = \frac{47.7}{200} \times 100 = 23.85%$

Final Answer: 23.85%

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JEE Advanced 2028 (Class 11)