91Ó°ÊÓ

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

A solution have $80gofNaN{O}_3$in $75gof{H}_{2}O$

Temperature is ${50}^{∘}C$and to ${20}^{∘}C$

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

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,

localid="1652685655746" $\frac{88\mathrm{g}{\mathrm{NaNO}}_3}{100g{H}_2\mathrm{O}}\text{and}\frac{100g{H}_2\mathrm{O}}{88\mathrm{g}{\mathrm{NaNO}}_3}$

Find out the grams of $NaN{O}_3$out of localid="1652685822899" $80gofNaN{O}_3$in $75gof{H}_{2}O$

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

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

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 $80gofNaN{O}_3$in $75gof{H}_{2}O$at${20}^{∘}C$

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

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

$$\begin{gathered} \text{crystallizable}{\mathrm{NaNO}}_3\text{at}{20}^{∘}\mathrm{C}=80.0\mathrm{g}−66\mathrm{g} \\ =14\mathrm{g} \end{gathered}$$$75gof{H}_{2}O$