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Current Question (ID: 20796)

Question:

$6.02 \times 10^{20}$ $\text{molecules of urea are present in } 100 \text{ mL of its solution.}$ $\text{The concentration of urea solution is:}$ $(\text{Avogadro constant, } N_A = 6.02 \times 10^{23} \text{ mol}^{-1})$

Options:

1. 0.001 \text{ M}
2. 0.01 \text{ M}
3. 0.02 \text{ M}
4. 0.1 \text{ M}

Solution:

$\text{Hint: Use the formula of molarity, that is, } \frac{\text{number of mole of urea}}{\text{volume of solution(L)}}$ $\text{Step 1:}$ $\text{The formula of molarity is as follows:}$ $\text{Molarity} = \frac{\text{number of mole of urea}}{\text{volume of solution(L)}}$ $6.02 \times 10^{20} \text{ molecules of urea is given}$ $6.022 \times 10^{23} \text{ molecules are present in 1 mole of urea.}$ $1 \text{ molecule urea is present in } \frac{1}{6.022 \times 10^{23}} \text{ moles of urea}$ $6.02 \times 10^{20} \text{ molecules of urea are present in } \frac{1}{6.022 \times 10^{23}} \times 6.02 \times 10^{20} \text{ moles of urea}$ $= 0.001 \text{ moles}$ $\text{The volume of solution is } 100 \text{ mL.}$ $\text{Step 2:}$ $\text{Calculate the molarity of the solution as follows:}$ $\text{Molarity} = \frac{0.001}{0.1}$ $\text{Molarity} = 0.01 \text{ M}$

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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
}