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Current Question (ID: 18051)

Question:
$\text{The density of } \text{H}_2\text{SO}_4 \text{ solution is } 1.84 \text{ g ml}^{-1}. \text{ In a 1 litre solution if } \text{H}_2\text{SO}_4 \text{ is } 93 \% \text{ by (w/v) then the molality of the solution is -}$
Options:
  • 1. $9.42 \text{ m}$
  • 2. $10.42 \text{ m}$
  • 3. $11.42 \text{ m}$
  • 4. $12.42 \text{ m}$
Solution:
$\text{Hint: molality} = \frac{\text{Moles}}{\text{wt. of water in kg}}$ $\text{Step 1:}$ $\text{Calculate the amount of solution as follows:}$ $\text{Given } \text{H}_2\text{SO}_4 \text{ is } 93\% \text{ by volume}$ $\text{Wt. of } \text{H}_2\text{SO}_4 = 93 \text{ g}$ $\text{Volume of solution} = 100 \text{ ml} \therefore \text{Density} = \frac{\text{mass}}{\text{volume}}$ $\therefore \text{mass} = d \times \text{volume}$ $\therefore \text{weight of solution} = 100 \times 1.84 \text{ g} = 184 \text{ g}$ $\text{weight of water} = 184 - 93 = 91 \text{ g}$ $\text{Step 2:}$ $\text{Calculate the molality of solution is as follows:}$ $\text{Molality} = \frac{\text{Moles}}{\text{wt. of water in kg}} = \frac{93 \times 1000}{98 \times 91} = 10.42$

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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
}