Import Question JSON

Current Question (ID: 21409)

Question:

$\text{A solution is prepared by dissolving } 0.6 \text{ g of urea (molar mass = } 60 \text{ g mol}^{-1}) \text{ and } 1.8 \text{ g of glucose (molar mass = } 180 \text{ g mol}^{-1}) \text{ in } 100 \text{ mL of water at } 27 ^\circ \text{C. The osmotic pressure of the solution is:}$ $\text{(R = } 0.08206 \text{ L atm K}^{-1}\text{mol}^{-1})$

Options:

1. $8.2 \text{ atm}$
2. $2.46 \text{ atm}$
3. $4.92 \text{ atm}$
4. $1.64 \text{ atm}$

Solution:

$\text{Hint: } \Pi_{\text{solution}} = \Pi_{\text{urea}} + \Pi_{\text{glucose}}$ $\text{Step 1:}$ $\text{When two solutions are fixed then their osmotic pressure gets added.}$ $\text{The formula used to calculate the osmotic pressure of a solution is as follows:}$ $\Pi_{\text{solution}} = \Pi_{\text{urea}} + \Pi_{\text{glucose}}$ $\Pi_{\text{solution}} = i_1 C_1 RT + i_2 C_2 RT$ $\text{Here, } i \text{ is Van't Hoff factor, and } C_1, C_2 \text{ are the concentration of urea}$ $\text{and glucose respectively.}$ $\text{Step 2:}$ $\text{Calculate the osmotic pressure of the solution as follows:}$ $\pi_{\text{solution}} = i_1 C_1 RT + i_2 C_2 RT$ $= (i_1 C_1 + i_2 C_2) RT$ $= \left( \frac{0.6 \times 1000}{60 \times 1000} + \frac{1.8 \times 1000}{180 \times 1000} \right) RT$ $= (0.1 + 0.1) RT$ $= 0.2 RT$ $= 0.2 \times 0.082 \times 300$ $= 4.92 \text{ atm}$

Import JSON File

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
}