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Current Question (ID: 10720)

Question:

$\text{On a new scale of temperature, which is linear and called the W scale, the freezing and boiling points of water are } 39^{\circ} \text{W and } 239^{\circ} \text{W respectively. What will be the temperature on the new scale corresponding to a temperature of } 39^{\circ} \text{C on the Celsius scale?}$

Options:

1. $78^{\circ} \text{C}$
2. $117^{\circ} \text{W}$
3. $200^{\circ} \text{W}$
4. $139^{\circ} \text{W}$

Solution:

\text{Whenever it is required to go from a higher fixed point to a lower fixed point, then we use the following method:} \frac{\text{(Temperature on known scale)} - \text{(LFP on known scale)}}{\text{(UFP)} - \text{(LFP)}} = \frac{\text{(Temperature on unknown scale)} - \text{(LFP on unknown scale)}}{\text{(UFP)} - \text{(LFP)}} \text{By putting the given values in the above equation, we get:} \frac{39-0}{100-0} = \frac{t-39}{239-39} \text{From the above relation, we get, } t = 117°\text{W}

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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
}