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Current Question (ID: 20778)

Question:

$0.4 \text{ g mixture of NaOH, Na}_2\text{CO}_3 \text{ and some inert impurities was first titrated with } \frac{N}{10} \text{ HCl using phenolphthalein as an indicator, } 17.5 \text{ mL of HCl was required at the end point. After this methyl orange was added and titrated. } 1.5 \text{ mL of same HCl was required for the next end point. The weight percentage of Na}_2\text{CO}_3 \text{ in the mixture is- (Rounded-off to the nearest integer)}$

Options:

1. $6 \%$
2. $4 \%$
3. $7 \%$
4. $9 \%$

Solution:

$\text{Hint: Use moles of HCl to find Na}_2\text{CO}_3 \text{ mass.}$ $\text{Upto first end point}$ $\text{gm equi. of (NaOH + Na}_2\text{CO}_3) = \text{HCl}$ $x + y \times 1 = \frac{1}{10} \times 17.5$ $x + y = 1.75 \quad \cdots (1)$ $\text{Upto second end point}$ $\text{NaOH + Na}_2\text{CO}_3 \equiv \text{HCl}$ $x + y \times 2 = \frac{1}{10} \times 19$ $x + 2y = 1.9$ $y = 0.15 \quad \cdots (2)$ $\% \text{Na}_2\text{CO}_3 = \frac{0.15 \times 10^{-3} \times 106}{0.4} \times 100$ $= 3.975\% = 4\%$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}