Import Question JSON

Current Question (ID: 8179)

Question:

$\text{A certain hydrate has the formula } \text{MgSO}_4 \cdot \text{x} \text{H}_2 \text{O}. \text{ A quantity of 54.2 g of the compound is heated in an oven to drive off the water. If the water vapour generated exerts a pressure of 24.8 atm in a 2.0 L container at } 120^{\circ} \text{C}, \text{ the value of x is-}$

Options:

1. $2$
2. $5$
3. $6$
4. $7$

Solution:

$\text{Hint: Ideal gas equation}$\n\n$\text{Step 1:}$\n\n$\text{The formula of ideal gas is}$\n\n$\text{PV = nRT , n = moles of water vapour}$\n\n$\text{Step 2:}$\n\n$\text{Calculate the number of moles of water vapour as follows:}$\n\n$n = \frac{PV}{RT} = \frac{24.8 \times 2}{0.0821 \times 393} = 1.53 \text{ moles}$\n\n$\text{The weight of water vapour = 1.53 } \times 18 = 27.6 \text{ g}$\n\n$\text{wt of } \text{MgSO}_4 = 54.2 - 27.6 = 26.5 \text{ g}$\n\n$\text{Step 3:}$\n\n$\text{Find the ratio of } \text{MgSO}_4 \text{ and } \text{H}_2\text{O.}$\n\n$\begin{array}{lccc} \text{Substance} & & \text{number of moles} & \text{Simple ratio} \\ 1. \text{ MgSO}_4 & = \frac{26.5}{120} & 0.22 & \frac{0.22}{0.22} = 1 \\ 2. \text{ H}_2\text{O} & = \frac{27.6}{18} & 1.53 & \frac{1.53}{0.22} \approx 7 \end{array}$

Import JSON File

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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
}