NTA Abhyas 2020
- Exam
- NTA Abhyas
- Year
- 2020
- Subject
- Chemistry
- Chapter
- Redox Reactions
NTA Abhyas 2020 · Chemistry · Redox Reactions
A 2.0 g sample of a mixture containing sodium carbonate, sodium...
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) HCl for complete neutralization. Calculate the percentage composition of in the original mixture.
Correct answer
31.50
Step-by-step solution
In 2g sample CO2 Liberated
Moles of CO2
moles = 5 mM of NaHCO3 in mixture
Moles of NaHCO3 in 2g sample = 2 x 5 x 10 mM
1.5g sample of HCl
2.0g sample
20 mM HCl will be utilised by
10 mM
{\text{NaHCO}}_{3} = \text{10 mM} = \text{10} \times 1 {0}^{- 3} \times 8 4 = \text{0.84g} \left(\text{42%}\right)
\left(\text{26.5%}\right){\text{Na}}_{2} {\text{SO}}_{4} = \text{(By difference)} = \text{0.63g} \left(\text{31.5%}\right)
Moles of CO2
moles = 5 mM of NaHCO3 in mixture
Moles of NaHCO3 in 2g sample = 2 x 5 x 10 mM
1.5g sample of HCl
2.0g sample
20 mM HCl will be utilised by
10 mM
{\text{NaHCO}}_{3} = \text{10 mM} = \text{10} \times 1 {0}^{- 3} \times 8 4 = \text{0.84g} \left(\text{42%}\right)
\left(\text{26.5%}\right){\text{Na}}_{2} {\text{SO}}_{4} = \text{(By difference)} = \text{0.63g} \left(\text{31.5%}\right)