SolveItClass 9 · NCERT

NCERT Solutions · Class 9 Science Exploring Mixtures and their Separation

25 questions · 21 still being checked

Pause and Ponder 5.3–5.4 (part 5 of 14)

  1. Exercise 5.3

    Vinegar, used as a food preservative and additive, contains 5\displaystyle 5 % v/v acetic acid. Glacial acetic acid is a liquid, i.e., 100\displaystyle 100% acetic acid. If you want to make vinegar from glacial acetic acid, how would you proceed? Grade 8\displaystyle 8 Curiosity Chapter 9\displaystyle 9 76\displaystyle 76 Exploration|Grade 9\displaystyle 9

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Dilute the glacial acetic acid $\displaystyle 20$ times with water — $\displaystyle 5$ mL of glacial acetic acid made up to $\displaystyle 100$ mL of solution gives $\displaystyle 5$ % v/v vinegar.
    NCERT_Solution_Class9_Science_Ch5_PP_Q5-3
    Vinegar must be $\displaystyle 5$ % v/v, i.e. \(\displaystyle \dfrac{\text{Volume of acetic acid}}{\text{Volume of solution}} \times 100 = 5 \).
    For $\displaystyle 100$ mL of vinegar: \(\displaystyle \text{Volume of acetic acid} = \dfrac{5 \times 100}{100} = 5\ \text{mL} \)
    For $\displaystyle 1$ litre of vinegar: \(\displaystyle \dfrac{5 \times 1000}{100} = 50\ \text{mL} \) of glacial acetic acid, water added up to the $\displaystyle 1000$ mL mark.
    Steps: take some water in a $\displaystyle 100$ mL measuring flask, add $\displaystyle 5$ mL of glacial acetic acid to it, then add water up to the $\displaystyle 100$ mL mark and stir.
    Safety first: glacial acetic acid is corrosive — measure and add it only under your teacher's supervision, and always add the acid to the water, never water to the acid.

  2. Exercise 5.4

    Filter the hot solution to remove insoluble impurities (Fig. 5.8b). Collect it in a clean beaker and cover it with a watch glass.NCERT_Question_Class9_Science_Ch5_PP_Q5-4

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Compound 'B' is likely to deposit more solid.
    The solid that separates on cooling = (solubility at $\displaystyle 80$ °C) − (solubility at $\displaystyle 60$ °C), per $\displaystyle 100$ g of water, so the compound whose curve falls more steeply between those two temperatures throws out more solid.
    Read Fig. $\displaystyle 5.6$ at the two temperatures: the curve of 'B' comes down through a large vertical distance from $\displaystyle 80$ °C to $\displaystyle 60$ °C, while the curve of 'A' is almost flat over the same interval, so the solubility of 'A' hardly changes and very little solid separates from it.
    How steep the curve of 'B' is can be seen from the one step the chapter puts numbers to — the $\displaystyle 60$ °C to $\displaystyle 40$ °C step of Section $\displaystyle 5.3.1$, where its solubility falls from $\displaystyle 287$ g to $\displaystyle 241$ g per $\displaystyle 100$ g of water, i.e. \(\displaystyle 287 - 241 = 46\ \text{g} \) of solid crystallising out of every $\displaystyle 100$ g of water. The $\displaystyle 80$ °C → $\displaystyle 60$ °C step lies higher up on that same steeply rising curve.
    Equal masses of the two hot saturated solutions are compared, so the answer turns only on how far each curve drops over $\displaystyle 80$ °C → $\displaystyle 60$ °C — and in Fig. $\displaystyle 5.6$ that drop is much the larger for 'B'.