Question

How many grams of commercial sodium hydroxide contains 75 percent $N{{a}_{2}}O$ are required to prepare 600 g of a 15 percent solution of $NaOH$?

Answer 1

Hint: The weight percent given in the two solutions are different and as per the question, the concentration or the weight of NaOH will be the same for both the conditions. Weight percent is a measurement of concentration of solution.

Complete step by step solution:
Solution can be prepared in the terms of relative percentage concentration of the solute in a solution. To determine the weight percent of a solution the mass of the solution i.e. the combined mass of solute and solvent and then it is multiplied by 100 to obtain it.
Given in the question:
Commercial sodium hydroxide contains 75 percent $N{{a}_{2}}O$
We have to find the amount of commercial sodium hydroxide which is required to prepare 600 g of a 15 percent solution of $NaOH$
Let x grams of $N{{a}_{2}}O$ is present in the solution
Then the pure $N{{a}_{2}}O$ present in the solution = $x(\dfrac{75}{100})$=$\dfrac{3x}{4}$
Amount of pure $NaOH$ present = $600(\dfrac{15}{100})$= 90 g
Molar mass of $NaOH$ = 40
The number of moles can be calculate by the ratio of given mass to the molar mass
Number of moles of NaOH = $\dfrac{90}{40}$= 2.25 moles
The reaction which is involved in the question :
\[N{{a}_{2}}O+{{H}_{2}}O\to 2NaO\]
1.125 moles of $N{{a}_{2}}O$ produce 2.25 moles of $NaOH$
= $\dfrac{3x}{(4)(62)}=1.125$

Hence the value of x = 93 g.

Note: Percentage weight by volume is a measurement of the concentration of the solution. Solubility is something that is given in the units of gram of solute per 100ml of water. Solutions can be described by other concentrations besides molarity, molality or normality.