Import Question JSON

Current Question (ID: 20780)

Question:

$\text{A 100 mL solution was made by adding 1.43 g of } \text{Na}_2\text{CO}_3\cdot x\text{H}_2\text{O}. \text{ The normality of the solution is 0.1 M. The value of } x \text{ is:}$ $\text{(The atomic mass of Na is 23 g/mol)}$

Options:

1. $1$
2. $10$
3. $4$
4. $7$

Solution:

$\text{Hint: } G_{\text{meq}} = \frac{\text{wt}}{\text{Eq}_{\text{wt}}}$ $\text{Molar mass of } \text{Na}_2\text{CO}_3\cdot x\text{H}_2\text{O}$ $= 23 \times 2 + 12 + 48 + 18x$ $= (106 + 18x)$ $\text{Eq}_{\text{wt}} = \frac{M}{2} = (53 + 9x)$ $\text{As } n_{\text{factor}} \text{ in dissolution will be determined from net cationic or anionic charge; which is 2 so}$ $\text{Eq}_{\text{wt}} = \frac{M}{2} = 53 + 9x$ $G_{\text{meq}} = \frac{\text{wt}}{\text{Eq}_{\text{wt}}} = \frac{1.43}{53 + 9x}$ $\text{Normality} = \frac{G_{\text{meq}}}{V_{\text{litre}}}$ $\text{Normality} = 0.1 = \frac{1.43}{53 + 9x} \cdot \frac{100}{0.1} \text{ ml}$ $\text{As volume} = 100 \text{ ml} = 0.1 \text{ Litre}$ $\text{So } 10^{-2} = \frac{1.43}{53 + 9x}$ $53 + 9x = 143$ $9x = 90$ $x = 10.00$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}