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Current Question (ID: 18083)

Question:
$\text{A solution containing 30 g of non-volatile solute in 90 g of water has a vapour}$ $\text{pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the}$ $\text{solution and the new vapour pressure becomes 2.9 kPa at 298 K.}$ $\text{The molar mass of the solute will be -}$
Options:
  • 1. $23 \text{ g mol}^{-1}$
  • 2. $34 \text{ g mol}^{-1}$
  • 3. $15 \text{ g mol}^{-1}$
  • 4. $46 \text{ g mol}^{-1}$
Solution:
$\text{Hint: Use relative lowering in vapour pressure formula.}$ $\text{Step 1:}$ $\text{The formula of relative lowering in vapour pressure is as follows:}$ $\frac{\Delta P}{P_1^0} = X_{\text{solute}}$ $\text{Calculate the mole of water (solvent) and solute (X) as follows:}$ $\text{Number of mole} = \frac{\text{amount of } X}{\text{molar mass of } X}$ $\frac{P_1^0 - 2.8}{P_1^0} = \frac{30}{30 + 5M}$ $1 - \frac{2.8}{P_1^0} = \frac{30}{30 + 5M}$ $\frac{5M}{30 + 5M} = \frac{2.8}{P_1^0} \quad \text{........(1)}$ $\text{Step 2:}$ $\text{18 g of water is then added to the solution. The moles of water changes. The}$ $\text{new moles of water is as follows:}$ $= \frac{90 + 18}{18} = 6 \text{ mol}$ $\frac{P_1^0 - P}{P_1^0} = \frac{30}{M} \times \frac{1}{30 + 6}$ $\frac{P_1^0 - 2.9}{P_1^0} = \frac{30}{M} \times \frac{1}{30 + 6M}$ $1 - \frac{2.9}{P_1^0} = \frac{30}{30 + 6M}$ $\frac{6M}{30 + 6M} = \frac{2.9}{P_1^0} \quad \text{.......(2)}$ $\text{Step 3:}$ $\text{Divide 1 and 2 as follows:}$ $\frac{\frac{5M}{30 + 5M}}{\frac{6M}{30 + 6M}} = \frac{\frac{2.8}{P_1^0}}{\frac{2.9}{P_1^0}}$ $\frac{5M}{30 + 5M} \times \frac{30 + 6M}{6M} = \frac{2.8}{2.9}$ $(30 + 6M) \times 5 \times 2.9 = 2.8 \times 6 \times (30 + 5M)$ $435 + 87M = 504 + 84M$ $87M - 84M = 504 - 435$ $3M = 69$ $M = \frac{69}{3}$ $M = 23u$ $\text{Hence, molar mass of X is 23 u}$

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{
  "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
}