Q. 9.99 - An Introduction to General, Organic, and Biological Chemistry

Question

What is the molarity of a solution containing $8.0g of NaOH$ in $400.mL of NaOH$ solution? $(9.4)$

Step-by-Step Solution

Answer

The $NaOH$ of the mobality is $0.500M$

Step 1: Given information

The molarity of a solution is described as the amount of solute per litre of solution. Compounds or solutions are two or more constituents systems that are homogenous. A mixture of treated water and some disintegrating material, such as a solid or gas; nevertheless, solutions of just one solid in an another are also common, such as gold and silver alloys. Any homogenous system with components is referred to as a solution (liquid, solid, or gas).

Step 2: Explanation.

$M$ represents it. A solution's molarity is defined as the number of moles of solutes per litre of solution. The following formula describes it:

$Molarity=\frac{moles of solute}{liter of solution}$

Step 3: substance's density

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

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

Step 4: Mass percentage

Mass percentage is a volume expression for a solute that could be estimated of the solution's mass. The following formula expresses it:

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

Step 5: moles of a solution

To find the moles of a solution containing $8.0g of NaOH$ in $400mL of NaOH$ solution, first work out just how many moles there are:

$No. of moles=mass of substitutes\times \frac{1mol}{molar mass}$

Step 6: Explanation.

$No.of moles of NaOH=mass of NaOH\times \frac{1mole of NaOH}{Molar mass of NaOH}$

$=8.0g\times \frac{1 mol NaOH}{39.997g}$
$=0.20mol NaOH$

Step 7: Conversion

Convert first $mL$ into $L$:

$1000mL=1.00L$

$400\cancel{mL}\times \frac{1.00L}{1000\cancel{mL}}=0.400mL$

Step 8: Explanation.

These measurements are now used to compute the molality of $NaOH$ using the above formula.

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

$\text{ Molárity }\left(\mathrm{M}\right)\text{ of }\mathrm{NaOH}=\frac{0.20\mathrm{molNaOH}}{0.400\mathrm{L}\text{ water }}$

$=0.500 mol/L$

$=0.500M$