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Current Question (ID: 7435)

Question:
The chloroform contamination level in water is 15 ppm (by mass of chloroform). The molality of chloroform in the water sample would be:
Options:
  • 1. $3.25 \times 10^{-4}$
  • 2. $1.5 \times 10^{-3}$
  • 3. $7.5 \times 10^{-3}$
  • 4. $1.25 \times 10^{-4}$
Solution:
\text{HINT: Use general expression of mass % and molality}\n\n\text{STEP 1: Calculate mass %}\n\n(i) 1 ppm \text{ is equivalent to 1 part out of 1 million } (10^6) \text{ parts.}\n\text{Mass percent of 15 ppm chloroform in water } = \frac{15}{10^6} \times 100 = 1.5 \times 10^{-3} \%\n\n\text{STEP 2: Calculate molality}\n\n(ii) 100 g \text{ of the sample contains } 1.5 \times 10^{-3} \text{ g of } \text{CHCl}_3.\n\Rightarrow 1000 \text{ g of the sample contains } 1.5 \times 10^{-2} \text{ g of } \text{CHCl}_3.\n\n\text{Molar mass of } \text{CHCl}_3 = 12.00 + 1.00 + 3(35.5)\n= 119.5 \text{ g mol}^{-1}\n\n\text{Molality } = \frac{15}{119.5} \times \frac{1000}{10^6}\n\n\text{Molality of chloroform in water } = 0.0125 \times 10^{-2} \text{ m}\n= 1.25 \times 10^{-4} \text{ m}

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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
}