Question

What is the maximum number of grams of \[N{H_4}Cl\] that will dissolve in \[200\] grams of water at \[70\] degree Celsius?

Answer 1

Hint: Remember that to dissolve is to enable a solute to completely enter into a solution; this is generally used in Chemistry. Dissolution is another word for dissolving. Dissolution usually means for the solid form to complete transition into a liquid form. Take the help of a solubility graph to further solve the question.

Complete answer:
Let us note down the given values;
The compound given to us is ammonium chloride that is \[N{H_4}Cl\].
The mass of water is given as \[200\] grams.
The temperature at which the process is to be carried out is \[70\] degree Celsius.
So what we need to do is find the maximum mass of ammonium chloride that can dissolve at the given values of temperature and mass of water. This means the mass that is completely soluble must be found.
We must take the help of a solubility graph in order to clearly recognize how much of a given salt dissolves at different temperatures. The graph we are using have the x-axis as the ‘temperature at degree Celsius’ and the y-axis gives ‘solute per \[100\] grams of water’.

Now since we are given the temperature at which the dissolution takes place, that is \[70\] degree Celsius, so we look at the x-axis of the graph where \[70\] degrees has been marked. Look at the y-axis for the value corresponding to \[70\] degrees on the x-axis. The y-axis value we find is approximately $62$ grams.
So we can infer in general that $62$ grams of solute per \[100\] grams of water gets dissolved at \[70\] degrees Celsius. This also means that at \[70\] degree Celsius, no more than $62$ grams can be dissolved in \[100\] grams of water. Any mass of salt added after $62$ grams will not be soluble in the water.
Now that we know the solubility of ammonium chloride in \[100\] grams of water, we are asked to find solubility in $200$ grams. This can be found using unitary method;
If \[100\] grams of water dissolves $62$ grams $ \Rightarrow 100g \to 62g$
$ \Rightarrow $1 gram of water would dissolve $ = \dfrac{{62}}{{100}}$ grams of ammonium chloride
Then $200$ grams of water would dissolve $ = ?$
$ \Rightarrow $Number of grams of ammonium chloride dissolved in \[200\] grams of water $ = 200 \times \dfrac{{62}}{{100}}$
$\therefore $Number of grams of ammonium chloride dissolved in \[200\] grams of water $ = 124g$ of $N{H_4}Cl$.

Note:
The dissolution of solute per 100 grams of water for each salt varies. To find the corresponding values of solubility, each salt can be plotted on the solubility curve, from there we can easily find the level of solubility of each salt at any given temperature.