Q. 9.87 - An Introduction to General, Organic, and Biological Chemistry

Question

Potassium nitrate has a solubility of $32 g of KN{O}_3 in 100. g of H_{2}O at 20°C$. Determine if each of the following forms an unsaturated or saturated solution at $20°C:(9.3)$

a. adding $32 g of KN{O}_3 to 200. g of H_{2}O$

b. adding $19 g of KN{O}_3 to 50. g of H_{2}O$

c. adding $68 g of KN{O}_3 to 150. g of H_{2}O$

Step-by-Step Solution

Answer

(Part a) $KN{O}_3$ was entirely dissolved in water, resulting in an unsaturated solution

(Part b) The leftover $3 g$ of solid remained undissolved, resulted in a saturated solution.

(Part c) the rest $20 g$ of solid is left undissolved, leading in a saturated solution.

Step 1: Introduction (Part a).

(a) At a specific temperature, a saturated mixture containing the maximum quantity of liquid sample. There is very little solute in an imbalanced solution than could be absorbed.

Step 2: Given data (a).

We must first establish the maximum quantity of $KN{O}_3$ that can dissolve in $200 g of H_{2}O at 2°C$ in order to ascertain whether an unsaturated or saturated solution occurs when $2 g of KN{O}_3$ is given to $200 g of H_{2}O at 2°C.$

Potassium nitrate has such a reactivity of $32 g of KN{O}_3 in 100 g of H_{2}O$ at a ratio of $20°C$. The relevant data can be gathered can be obtained using the soluble data:

$\frac{32 g KN{O}_3}{100 g H_{2}O} and \frac{100 g H_{2}O}{32 g KN{O}_3}$

Step 3: Explanation (a).

The greatest quantity of $KN{O}_3$ that may be absorbed in $200 g of H_{2}O$ can now be computed as follows:

$200. \cancel{g H_{2}O}\times \frac{32 g KN{O}_3}{100 \cancel{g H_{2}O}}=64 g KN{O}_3$

The highest quantity of $KN{O}_3$ that may dissolve in $200 g of H_{2}O at 20°C is 64 g.$. When $2 g of KN{O}_3$ is mixed with $200 g$ of water, it totally dissolves and creates an unsaturated solution.

Step 4: Given data (b).

(b) Specify the total quantity of $KN{O}_3$ that can breakdown in $50 g of H_{2}O$to identify if an unsaturated or saturated solution occurs when $19 g of KN{O}_3$ is given to $50 g of H_{2}O at 20°C$.

Potassium nitrate does have a concentration of $32 g of KN{O}_3 in 100 g$ of water at $20°C$. The appropriate data obtained can be derived using the solubility data:

$\frac{32 g KN{O}_3}{100 g H_{2}O} and \frac{100 g H_{2}O}{32 g KN{O}_3}$

Step 5: Explanation (b).

The largest quantity of $KN{O}_3$ that may melt in $50 g of H_{2}O$ can now be computed as follows:

$50. \cancel{g H_{2}O}\times \frac{32 g KN{O}_3}{100 \cancel{g H_{2}O}}=16 g KN{O}_3$

The quantity of $KN{O}_3$ that may absorb in $50 g of H_{2}O at 20°C is 16 g.$ Only $19 g of KN{O}_3$ can mix fully in water if $16 g of KN{O}_3$ is introduced to$50 g$ of water, remaining $3 g$ of material undissolved to produce a saturated solution.

Step 6: Given data (c).

(c) Specify the maximum quantity of $KN{O}_3$ that can breakdown in $150 g of H_{2}O$ to establish how an unsaturated or saturated solution develops when $68 g of KN{O}_3$ is added to $150 g of H_{2}O at 20°C$.

Potassium nitrate really does have a concentration of $32 g of KN{O}_3 in 100 g$  of water at $20°C$. The corresponding changes took place can be estimated using the solubility data:$\frac{32 g KN{O}_3}{100 g H_{2}O} and \frac{100 g H_{2}O}{32 g KN{O}_3}$

Step 7: Explanation (c).

The optimum quantity of $KN{O}_3$ that may melt in $150 g of H_{2}O$ can now be computed by dividing:

$150. \cancel{g H_{2}O}\times \frac{32 g KN{O}_3}{100 \cancel{g H_{2}O}}=48 g KN{O}_3$

The volume of $KN{O}_3$ that may disintegrate in $150 g of H_{2}O at 20°C is 48 g.$ Only $68 g of KN{O}_3$ can dissolve totally in water if $48 g of KN{O}_3$ is put to$150 g$of water, allowing $20 g$ of substance undissolved to create a saturated solution.