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Calculate the amount of solid NaOH in the initial 0.1 M NaOH solution portion that was diluted.

Chemistry

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Question c (from previous problem): Calculate the amount of solid NaOH in the initial 0.1 M NaOH solution portion that was diluted.

Given:

  • Molarity of NaOH solution (MM) = 0.1 M0.1 \text{ M}
  • Volume of NaOH solution (VV) = 5.0 cm35.0 \text{ cm}^3
  • Molar mass of NaOH: Na=23\text{Na} = 23, O=16\text{O} = 16, H=1\text{H} = 1
  1. Convert the volume from cm3\text{cm}^3 to Liters.

    V=5.0 cm3=5.0 mL=0.005 LV = 5.0 \text{ cm}^3 = 5.0 \text{ mL} = 0.005 \text{ L}
  2. Calculate the number of moles of NaOH.

    Moles=Molarity×Volume (in L)\text{Moles} = \text{Molarity} \times \text{Volume (in L)} Moles of NaOH=0.1 mol/L×0.005 L\text{Moles of NaOH} = 0.1 \text{ mol/L} \times 0.005 \text{ L} Moles of NaOH=0.0005 mol\text{Moles of NaOH} = 0.0005 \text{ mol}
  3. Calculate the molar mass of NaOH.

    Molar mass of NaOH=Na+O+H\text{Molar mass of NaOH} = \text{Na} + \text{O} + \text{H} Molar mass of NaOH=23+16+1=40 g/mol\text{Molar mass of NaOH} = 23 + 16 + 1 = 40 \text{ g/mol}
  4. Calculate the mass of NaOH.

    Mass=Moles×Molar mass\text{Mass} = \text{Moles} \times \text{Molar mass} Mass of NaOH=0.0005 mol×40 g/mol\text{Mass of NaOH} = 0.0005 \text{ mol} \times 40 \text{ g/mol} Mass of NaOH=0.02 g\text{Mass of NaOH} = \boxed{0.02 \text{ g}}

    Question 4: Study the diagrams below carefully

    a) Identify the types of crops illustrated in A, B and C.

    • A: Cocoyam
    • B: Cassava
    • C: Sweet potato

    b) What type of storage organ does each of the crops A, B and C represent?

    • A: Corm (modified stem)
    • B: Tuberous root (modified adventitious root)
    • C: Tuberous root (modified adventitious root)

    c) Name the parts labeled I, II, III, IV, V and VI.

    • I: Scaly leaves (or leaf scars)
    • II: Cormel (or sucker)
    • III: Main corm
    • IV: Fibrous roots
    • V: Tuberous root (or storage root)
    • VI: Adventitious roots (or root hairs)

    d) Tabulate three structural differences between the crops A and C.

    FeatureCrop A (Cocoyam)Crop C (Sweet potato)
    Type of organModified stem (corm)Modified root (tuberous root)
    External featuresHas scaly leaves/leaf scars and buds ("eyes")Has adventitious roots/root hairs; lacks scaly leaves
    Growth patternGrows as a compact, often rounded, underground stemGrows as an elongated, irregular swollen root

    e) State the usual methods of propagation of each of the crops in B and C.

    • B (Cassava): Stem cuttings
    • C (Sweet potato): Stem cuttings (slips) or sprouts from the tuberous roots
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