Q. 9.28 - An Introduction to General, Organic, and Biological Chemistry

Question

A solution containing $80 g of NaN{O}_3$ in $75 g of H_{2}O$ at ${50}^{\circ }C$ is cooled to ${20}^{\circ }C$ 

a. How many grams of $NaN{O}_3$ remain in solution at ${20}^{\circ }C$?

b. How many grams of solid $NaN{O}_3$ crystallized after cooling?

           

Step-by-Step Solution

Answer

(part a) $66 g of {\mathrm{NaNO}}_3$ remaining in the solution ${20}^{\circ }C$ when a solution containing $80 g of NaN{O}_3$ in $75 g of H_{2}O.$

(part b) $14 g of {\mathrm{NaNO}}_3$ remained in the solution crystalized after cooling.

Step1: Introduction (part a).

A solution in science is a blend of different substances in varying proportions that can be varied continuously up to the limit of solubilization.

Step2: Given Information (part a, b).

A solution have $80 g of NaN{O}_3$ in $75 g of H_{2}O$

Temperature is ${50}^{\circ }C$ and to ${20}^{\circ }C$

Step3: Explanation (part a).

Find the quantity of $NaN{O}_3$ in solution at ${20}^{\circ }C.$

Then,

$88\mathrm{g}\text{ of }{\mathrm{NaNO}}_3=100\mathrm{g}\text{ of }{\mathrm{H}}_2\mathrm{O}$

Therefore, the factor of conversion are,

$\frac{88\mathrm{g} {\mathrm{NaNO}}_3}{100g H_2\mathrm{O}}\text{ and }\frac{100g H_2\mathrm{O}}{88\mathrm{g} {\mathrm{NaNO}}_3}$

Find out the grams of $NaN{O}_3$ out of $80 g of NaN{O}_3$ in $75 g of H_{2}O$

$$\begin{gathered} \text{ Amount of }{\mathrm{NaNO}}_3=75\mathrm{g} {\mathrm{H}}_2\mathrm{O}\times \frac{80\mathrm{g} {\mathrm{NaNO}}_3}{100\mathrm{g} {\mathrm{H}}_2\mathrm{O}} \\ =66 g of {\mathrm{NaNO}}_3 \end{gathered}$$

Step4: Explanation (part b).

Find the quantity of $NaN{O}_3$ can be crystalized after cooling process at ${20}^{\circ }C$

Therefore,

$88\mathrm{g}\text{ of }{\mathrm{NaNO}}_3=100\mathrm{g}\text{ of }{\mathrm{H}}_2\mathrm{O}$

Hence, the factor of conversion are,

$\frac{88\mathrm{g} {\mathrm{NaNO}}_3}{100g H_2\mathrm{O}}\text{ and }\frac{100g H_2\mathrm{O}}{88\mathrm{g} {\mathrm{NaNO}}_3}$

Now find the grams of $NaN{O}_3$ in which $80 g of NaN{O}_3$in $75 g of H_{2}O$ at${20}^{\circ }C$

$$\begin{gathered} \text{ Amount of }{\mathrm{NaNO}}_3=75\mathrm{g} {\mathrm{H}}_2\mathrm{O}\times \frac{80\mathrm{g} {\mathrm{NaNO}}_3}{100\mathrm{g} {\mathrm{H}}_2\mathrm{O}} \\ =66 g of {\mathrm{NaNO}}_3 \end{gathered}$$

The grams of crystalized $NaN{O}_3$ is find by subtracting fully dissolved $NaN{O}_3$ at ${20}^{\circ }C$ with fully dissolved $NaN{O}_3$ at ${50}^{\circ }C,$ 

$$\begin{gathered} \text{ crystallizable }{\mathrm{NaNO}}_3\text{ at }{20}^{\circ }\mathrm{C}=80.0\mathrm{g}-66\mathrm{g} \\ =14 \mathrm{g} \end{gathered}$$$75 g of H_{2}O$