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Current Question (ID: 18106)

Question:

$\text{A solution of sucrose (molar mass = 342 g mol}^{-1}\text{) has been prepared by}$ $\text{dissolving 68.5 g of sucrose in 1000 g of water.}$ $\text{The freezing point of the solution obtained will be:}$ $\left( k_f \text{ for water = 1.86 K kg mol}^{-1} \right)$

Options:

1. $-0.372\ ^\circ \text{C}$
2. $0.372\ ^\circ \text{C}$
3. $0.572\ ^\circ \text{C}$
4. $-0.572\ ^\circ \text{C}$

Solution:

$\text{Hint: } \Delta T_f = k_f \times m$ $\text{Step 1:}$ $\text{Depression in freezing point formula is as follows:}$ $\Delta T_f = k_f \times m$ $\text{Calculate the value of molality as follows:}$ $\text{where, } m = \text{molality} = \frac{w(\text{solute})}{M(\text{molar mass})} \times \text{amount of solvent (kg)}$ $W_A = \frac{68.5}{342 \times 1} = 0.198\ m$ $\text{Step 2:}$ $\text{Calculate the temperature of solution as follows:}$ $\Delta T_f = 1.86 \times 0.198 = 0.372\ ^\circ \text{C}$ $\Delta T_f = T^\circ - T_s = 0 - 0.372\ ^\circ \text{C}$

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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
}