Q. 9.97 - An Introduction to General, Organic, and Biological Chemistry

Question

How many liters of a $12\% (m/v) KOH$ solution would you need to obtain $86.0 g of KOH$ ? $(9.4)$

Step-by-Step Solution

Answer

To acquire $86.0 g of KOH$, $0.716 L$ of a $12\% of KOH$ solution needed necessary.

Step 1: Introduction

It has a variety of industrial and niche applications, the majority of which take advantage of its caustic nature and acid reactivity.

Step 2: Given Data.

The molar mass of a solution is defined as the molar concentration per litre of solution. $M$ indicates it. A solution's molarity is described as the amount of units of solute molecules per litre of solution. The following formula explains it:

$Molarity\left(M\right)=\frac{moles of solutes}{liter of solution}$

Step 3: Explanation.

A substance's density is simply the ratio of its weight to its volume. A substance's density can be measured as its grams per cubic centimeter.

$\text{ density }\left(\mathrm{g}/{\mathrm{cm}}^3\right)=\frac{\mathrm{mass} \left(\mathrm{g}\right)}{\text{ volume }\left({\mathrm{cm}}^3\right)}\text{ or }\frac{\text{ mass }\left(\mathrm{g}\right)}{\text{ volume }\left(\mathrm{mL}\right)}$

Step 4: Given data.

The mass percent calculator will help decide the mass of an element in a substance in proportion to the overall amount of substance.

Step 5: Explanation.

Mass percentage is a concentration expression for a solute that can be estimated of the optimizer mass. The following formula defines it:

$mass percentage of solute=\frac{grams of solute}{grams of solution}\times 100\%$

Step 6: Given data.

The weight of the solute present provides the mass percentage (m/v).

$Mass percent in m/v =\left(\frac{mass of solute}{volume of solution}\right)\times 100\%$

Step 7: Explanation.

$Mass percent in m/v =\left(\frac{mass of solute}{volume of solution}\right)\times 100\%$

$12\%=\left(\frac{86.0g}{volume of solution in mL}\right)\times 100\%$

$Volume of solution in mL=\frac{86}{12}\times 100$

$=716mL or 0.716L$

To produce $86.0 g of KOH$, $0.716 L$ of a $12\% of KOH$ solution was required.