Import Question JSON

Current Question (ID: 20782)

Question:

$\text{A solution of two components containing } n_1 \text{ moles of the 1}^{\text{st}} \text{ component } n_2 \text{ moles of the 2}^{\text{nd}} \text{ component is prepared.}$ $M_1 \text{ and } M_2 \text{ are the molecular weights of component 1 and 2 respectively.}$ $\text{If } d \text{ is the density of the solution in g mL}^{-1}, C_2 \text{ is the molarity and } x_2 \text{ is the mole fraction of the 2}^{\text{nd}} \text{ component, then } C_2 \text{ can be expressed as:}$

Options:

1. $C_2 = \frac{dx_2}{M_2 + x_2 \left( M_2 - M_1 \right)}$
2. $C_2 = \frac{1000 \ dx_2}{M_1 + x_2 \left( M_2 - M_1 \right)}$
3. $C_2 = \frac{1000x_2}{M_1 + x_2 \left( M_2 - M_1 \right)}$
4. $C_2 = \frac{dx_1}{M_2 + x_2 \left( M_2 - M_1 \right)}$

Solution:

$\text{Hint: Volume of solution} = \frac{(x_2M_2 + (1-x_2)M_1)}{d} \text{ mL}$ $\text{Explanation:}$ $\text{Given}$ $\text{For 1}^{\text{st}} \text{ component -}$ $\text{number of moles} = n_1 \text{ molecular weight} = M_1$ $\text{For the 2}^{\text{nd}} \text{ component (solute) - number of moles} = n_2,$ $\text{molecular weight} = M_2 \text{ mole fraction} = x_2$ $\text{2. Calculation of molarity of 2}^{\text{nd}} \text{ component } (C_2):$ $\text{Molarity} = \frac{\text{number of moles of solute}}{\text{volume of solution mL}} \times 1000$ $\text{Step 1: Calculation of moles 2}^{\text{nd}} \text{ component -}$ $\text{Let the total number of moles present in the solution be 1.}$ $n_1 + n_2 = 1$ $\text{We know, } x_2 = \frac{n_2}{n_1 + n_2}$ $\text{So, } x_2 = n_2 \text{ (number of moles of solute/2}^{\text{nd}} \text{ component)}$ $\text{and } 1-x_2 = n_1$ $\text{Step 2: Calculation of mass of solution and solute:}$ $\text{Number of moles of solute/component} = x_2$ $\text{So, mass} = x_2M_2g$ $\text{Mass of 1}^{\text{st}} \text{ component} = (1-x_2)M_1g$ $\text{Mass of the solution} = (x_2M_2 + (1-x_2)M_1)g$ $\text{Step 3: Calculation of volume of the solution:}$ $\text{Volume} = \frac{\text{mass}}{\text{density}}$ $\text{Volume of solution} = \frac{(x_2M_2 + (1-x_2)M_1)}{d} \text{ mL}$ $\text{Step 4: Calculation of molarity of 2}^{\text{nd}} \text{ component } (C_2) \text{ - Putting the values of the number of moles of solute and volume of solution in equation 1, we get:}$ $C_2 = \frac{1000 \ d \ x_2}{M_1 + x_2(M_2 - M_1)}$

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
}