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Current Question (ID: 10439)

Question:

$\text{Three liquids of densities } d, 2d \text{ and } 3d \text{ are mixed in equal proportions of weights. The relative density of the mixture is:}$

Options:

1. $\frac{11d}{7}$
2. $\frac{18d}{11}$
3. $\frac{13d}{9}$
4. $\frac{23d}{18}$

Solution:

\text{Hint: } \rho_{\text{mix}} = \frac{M_{\text{mix}}}{V_{\text{mix}}} \text{Step 1: Calculate the relative density of the mixture.} \text{The density of the three liquids are: } d, 2d, 3d \text{Let the mass of the liquids be } m \text{So the volumes of the three liquids are: } \frac{m}{d}, \frac{m}{2d}, \frac{m}{3d} \text{So, the mass of the mixture is: } M_{\text{mix}} = 3m \text{ ... (1)} \text{The volume of the mixture is given by:} V_{\text{mix}} = \frac{m}{d} + \frac{m}{2d} + \frac{m}{3d} = \frac{11m}{6d} \text{ ... (2)} \text{The relative density of the mixture is given by:} \rho_{\text{mix}} = \frac{M_{\text{mix}}}{V_{\text{mix}}} = \frac{3m}{\frac{11m}{6d}} = \frac{18d}{11} \text{Hence, option (2) is the correct answer.}

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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
}