Question

What is the molarity of a solution containing $8.0gofNaOH$in $400.mLofNaOH$solution? $(9.4)$

Short answer

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

Step-by-step solution

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).

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:

$Molarity=\frac{molesofsolute}{literofsolution}$

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.

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

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:

$Masspercentageofsolute=\frac{gramsofsolute}{gramsofsolution}\times 100\%$

moles of a solution 

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

$No.ofmoles=massofsubstitutes\times \frac{1mol}{molarmass}$

Explanation.

$No.ofmolesofNaOH=massofNaOH\times \frac{1moleofNaOH}{MolarmassofNaOH}$

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

Conversion

Convert first $mL$into $L$:

$1000mL=1.00L$

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

Explanation.

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

$Molarity\left(M\right)=\frac{molesofsolutes}{literofsolution}$

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

$=0.500mol/L$

$=0.500M$