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3.23 The solubility of
Cr(NO_(3))_(3)*9H_(2)O in water is
208g per
100_(g) of water at
15^(@)C. A solution of
Cr(NO_(3))_(3)*9H_(2)O in water at
35^(@)C is formed by dissolving
324g in
100g of water. When this solution is slowly cooled to
15^(@)C, no precipitate forms, (a) Is the solution that has cooled down to
15^(@)C unsaturated, saturated, or supersaturated? (b) You take a metal spatula and scratch the side of the glass vessel that contains this cooled solution, and crystals start to appear. What has just happened? (c) At equilibrium, what mass of crystals do you expect to form?

$3$ . $23$ The solubility of $\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} \cdot 9 \mathrm{H}_{2} \mathrm{O}$ in water is $208 \mathrm{~g}$ per $100_{\mathrm{g}}$ of water at $15^{\circ} \mathrm{C}$ . A solution of $\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} \cdot 9 \mathrm{H}_{2} \mathrm{O}$ in water at $35^{\circ} \mathrm{C}$ is formed by dissolving $324 \mathrm{~g}$ in $100 \mathrm{~g}$ of water. When this solution is slowly cooled to $15^{\circ} \mathrm{C}$ , no precipitate forms, (a) Is the solution that has cooled down to $15^{\circ} \mathrm{C}$ unsaturated, saturated, or supersaturated? (b) You take a metal spatula and scratch the side of the glass vessel that contains this cooled solution, and crystals start to appear. What has just happened? (c) At equilibrium, what mass of crystals do you expect to form?

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Volume of cubes and rectangular prisms: word problems

Full solution

Q.

3

.

23

The solubility of

\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} \cdot 9 \mathrm{H}_{2} \mathrm{O}

in water is

208 \mathrm{~g}

per

100_{\mathrm{g}}

of water at

15^{\circ} \mathrm{C}

. A solution of

\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} \cdot 9 \mathrm{H}_{2} \mathrm{O}

in water at

35^{\circ} \mathrm{C}

is formed by dissolving

324 \mathrm{~g}

in

100 \mathrm{~g}

of water. When this solution is slowly cooled to

15^{\circ} \mathrm{C}

, no precipitate forms, (a) Is the solution that has cooled down to

15^{\circ} \mathrm{C}

unsaturated, saturated, or supersaturated? (b) You take a metal spatula and scratch the side of the glass vessel that contains this cooled solution, and crystals start to appear. What has just happened? (c) At equilibrium, what mass of crystals do you expect to form?

Given Solubility Data: Solubility at $15^\circ\text{C}$ is $208\,\text{g}$ per $100\,\text{g}$ of water. The solution has $324\,\text{g}$ of $\text{Cr(NO}_3)_3\cdot9\text{H}_2\text{O}$ in $100\,\text{g}$ of water at $35^\circ\text{C}$ .
Exceeding Solubility Limit: When cooled to $15\,^\circ\mathrm{C}$ , the solubility should be $208\,\mathrm{g}$ , but we have $324\,\mathrm{g}$ in $100\,\mathrm{g}$ of water, which is more than the solubility limit.
Supersaturation Identified: Since the amount of solute is greater than the solubility limit at $15^\circ\text{C}$ , the solution is supersaturated.
Nucleation Initiated: Scratching the glass vessel likely initiated nucleation, causing the supersaturated solution to begin crystallizing.
Calculate Expected Crystal Mass: To find the mass of crystals expected to form, subtract the solubility limit at $15^\circ\text{C}$ from the initial amount of solute.
Subtracting Solubility Limit: Initial amount of solute is $324\text{g}$ . Solubility limit at $15^\circ\text{C}$ is $208\text{g}$ . Expected mass of crystals = $324\text{g} - 208\text{g}$ .
Final Crystal Mass Calculation: Expected mass of crystals = $324\text{g} - 208\text{g} = 116\text{g}$ .

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