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Current Question (ID: 8676)

Question:

$\text{Hardness of water cannot be removed by -}$

Options:

1. $\text{Clark's method}$
2. $\text{Adding washing soda}$
3. $\text{Calgon process}$
4. $\text{Filtration}$

Solution:

$\text{Hint: filtration is a physical process}$ $\text{Step 1:}$ $\text{HCO}_3^- = \text{temporary hardness}$ $\text{Cl}^- \text{ and SO}_4^{2-} = \text{Permanent hardness}$ $\text{Clark's method:}$ $\text{In this method, the calculated amount of lime is added to hard water. It precipitates out calcium carbonate and magnesium hydroxide which can be filtered off. The reaction is as follows:}$ $\text{Ca(OH)}_2 + \text{Ca(HCO}_3\text{)}_2 \rightarrow \text{CaCO}_3 \downarrow + \text{H}_2\text{O}$ $\text{Addition of washing soda}$ $\text{Washing soda reacts with soluble calcium and magnesium chlorides and sulphates in hard water to form insoluble carbonates. The reaction is as follows:}$ $\text{NaCO}_3 + \text{CaCl}_2 \rightarrow \text{CaCO}_3 \downarrow + \text{CaSO}_4$ $\text{Calgon Process}$ $\text{Sodium hexametaphosphate, Na}_6\text{(PO}_3\text{)}_6 \text{ commercially called 'calgon', is when added to hard water.}$ $\text{Na}_6\text{(PO}_3\text{)}_6$ $\text{Na}_2[\text{Na}_4\text{(PO}_3\text{)}_6]$ $\text{Ca}^{+2} \rightarrow \text{Na}_2[\text{CaNa}_2\text{(PO}_3\text{)}_6]$ $\text{Mg}^{+2} \rightarrow [\text{MgNa}_2\text{(PO}_3\text{)}_6]$ $\text{Filtration:}$ $\text{The physical process like filtration cannot use to remove hardness of water}$

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
}