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Amount of solute or solvent for a given elevation of boiling point

This type of question, asked 5 times: May 2022 S1, 22 Apr 2025 S1, 22 Apr 2025 S2, 25 Apr 2025 S2, 13 May 2026 S2. All 5 questions below, oldest first.

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  1. Question 1

    May 2022 S1

    97
    Calculate the amount of solute dissolved in 160-gram solvent that boils at 85°C, the molar mass of solute is 120 g mol−1^{-1}. KbK_b for solvent = 2.7°C kg mol−1^{-1} and boiling point of solvent = 76°C

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  2. Question 2

    22 Apr 2025 S1

    73
    Calculate the number of moles of nonvolatile solute dissolved in 0.3 kg solvent if ΔTb=0.3 K\Delta T_{b}= 0.3\text{ }K and KbK_{b} for solvent is 1.8 K kg mol−11.8\text{ }K\text{ }kg{\text{ }mol}^{- 1}.

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  3. Question 3

    22 Apr 2025 S2

    96
    When 0.01 mole of nonvolatile solute is dissolved in certain solvent calculate the mass of solvent in kg if ΔTb=0.6 K\Delta T_{b}= 0.6\text{ }K and KbK_{b} for solvent = 2 K kg mol−12\text{ }K\text{ }kg{\text{ }mol}^{- 1}

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  4. Question 4

    25 Apr 2025 S2

    67
    Calculate the number of moles of nonvolatile solute dissolved in 0.5 kg solvent if molal elevation constant for solvent is 2 kg K mol−1[ΔTb=0.8 K]2\text{ }kg\text{ }K{\text{ }mol}^{- 1}\left\lbrack \Delta T_{b}= 0.8\text{ }K \right\rbrack

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  5. Question 5

    13 May 2026 S2

    24
    Calculate the mass of solute dissolved in 100100 g solvent that produces boiling point elevation 0.50.5 K. [Kb[K_b for solvent =2 K kg mol−1= 2\ \mathrm{K\,kg\,mol^{-1}} & Molar mass of solute =120 g mol−1]= 120\ \mathrm{g\,mol^{-1}}]

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