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Current Question (ID: 21415)

Question:
$5\% \ (w/v) \ \text{solution of cane sugar (molar mass 342) is isotonic with} \ 1\% \ (w/v) \ \text{of a solution of an unknown solute. The molar mass of an unknown solute in g/mol is:}$
Options:
  • 1. $171.2$
  • 2. $68.4$
  • 3. $34.2$
  • 4. $136.2$
Solution:
$\text{Since it is given that the solutions are isotonic,}$ $\text{Here } \pi_1 = \text{for cane sugar solution}$ $\pi_2 = \text{for substance } x$ $\text{So } \pi_1 = \pi_2$ $\therefore \ C_1RT = C_2RT \quad \text{Where } C = \text{Concentration}$ $\Rightarrow \left( \frac{\text{wt. in gm.}}{\text{M. wt. of cane sugar}} \right) = \left( \frac{\text{wt. in gm.}}{\text{M. wt. of } X} \right)$ $\pi_1 = \pi_2$ $c_1 = c_2$ $\frac{5}{342} = \frac{1}{M}$ $M = \frac{342}{5} = 68.4$

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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
}