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How many potassium ions are present in 25 cm3 of 2M KOH? (Avogadro number = 6.02 x 1023)

Chemistry

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Answer

D

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Identify the gas X

The diagram shows a gas being collected in an inverted test tube, likely over water. This method is typically used for gases that are insoluble or sparingly soluble in water. Given the common context of such diagrams in chemistry, particularly when showing an electrode setup (as suggested by the partial image above), this often represents the collection of hydrogen gas from the electrolysis of water. Hydrogen gas (H2_2) is sparingly soluble in water and less dense than air, making this collection method suitable.

The correct option is D.

25. How many potassium ions are present in 25 cm3^3 of 2M KOH? (Avogadro number = 6.02 x 1023^{23})

  1. Convert the volume from cm3^3 to liters.

    V=25 cm3=25 mL=0.025 LV = 25 \text{ cm}^3 = 25 \text{ mL} = 0.025 \text{ L}

  2. Calculate the number of moles of KOH.

    Molarity (M) is defined as moles per liter.

    Moles of KOH=Molarity×Volume (L)\text{Moles of KOH} = \text{Molarity} \times \text{Volume (L)} Moles of KOH=2 M×0.025 L=0.05 mol\text{Moles of KOH} = 2 \text{ M} \times 0.025 \text{ L} = 0.05 \text{ mol}

  3. Determine the number of moles of potassium ions (K+^+).

    Potassium hydroxide (KOH) is a strong electrolyte and dissociates completely in water:

    KOH (aq)→K+(aq)+OH−(aq)\text{KOH (aq)} \rightarrow \text{K}^+ \text{(aq)} + \text{OH}^- \text{(aq)}

    For every 1 mole of KOH, 1 mole of K+^+ ions is produced.

    Moles of K+=0.05 mol\text{Moles of K}^+ = 0.05 \text{ mol}

  4. Calculate the total number of potassium ions.

    Number of ions=Moles×Avogadro’s number\text{Number of ions} = \text{Moles} \times \text{Avogadro's number} Number of K+ ions=0.05 mol×(6.02×1023 ions/mol)\text{Number of K}^+ \text{ ions} = 0.05 \text{ mol} \times (6.02 \times 10^{23} \text{ ions/mol}) Number of K+ ions=0.301×1023 ions\text{Number of K}^+ \text{ ions} = 0.301 \times 10^{23} \text{ ions} Number of K+ ions=3.01×1022 ions\text{Number of K}^+ \text{ ions} = 3.01 \times 10^{22} \text{ ions}

    The correct option is B.

    26. A gas of volume 420 cm3^3 has a pressure of 790 mmHg. What will be its new volume if the pressure reduces to 740 mmHg?

    This problem can be solved using Boyle's Law, which states that for a fixed amount of gas at constant temperature, the pressure and volume are inversely proportional (P1V1=P2V2P_1V_1 = P_2V_2).

  5. Identify the given values.

    Initial volume (V1V_1) = 420 cm3420 \text{ cm}^3
    Initial pressure (P1P_1) = 790 mmHg790 \text{ mmHg}
    Final pressure (P2P_2) = 740 mmHg740 \text{ mmHg}
    Final volume (V2V_2) = ?

  6. Apply Boyle's Law formula.

    P1V1=P2V2P_1V_1 = P_2V_2

  7. Solve for V2V_2.

    V2=P1V1P2V_2 = \frac{P_1V_1}{P_2} V2=(790 mmHg)×(420 cm3)740 mmHgV_2 = \frac{(790 \text{ mmHg}) \times (420 \text{ cm}^3)}{740 \text{ mmHg}} V2=331800740 cm3V_2 = \frac{331800}{740} \text{ cm}^3 V2≈448.378 cm3V_2 \approx 448.378 \text{ cm}^3

    Rounding to the nearest whole number, the new volume is approximately 448 cm3448 \text{ cm}^3.

    The correct option is C.

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