Question

How many grams of $NaOH$are needed to prepare $250\,mL$of $0.205\,M\,\,NaOH?$
(i) $3.10\,g$
(ii) $2.65\,g$
(iii) $2.05\,g$
(iv) $1.85\,g$

Answer 1

$M$ generally refers to the molarity of a solution which can be given by, $Molarity\, = \,\dfrac{{No.\,of\,moles\,of\,solute}}{{Volume\,of\,solution\,\left( {in\,L} \right)}}$. In the question molarity of the solution and the volume of solution is given from there calculate the number of moles of$NaOH$and hence find the mass of $NaOH$needed.

Complete step-by-step answer: The symbol $M$ mentioned in the question refers to molarity. Molarity is defined as the number of moles of solute per $L$ of solution.
Hence the molarity can be expressed as,
$Molarity\,\, = \,\,\dfrac{{No.\,of\,moles\,of\,solute}}{{Volume\,of\,solution\,\left( {in\,L} \right)}}............\left( 1 \right)$
Now, we are provided with a $NaOH$solution where the solute is $NaOH$.
The volume of the solution $= \,250\,mL$
Since the volume should be in $L$ we need to convert the given volume from $mL$ to $L$.
We know, $1\,L\, = \,1000\,mL$
Therefore, the volume of the solution of$NaOH\, = \,250\,mL \times \dfrac{{1\,L}}{{1000\,mL}}\, = \,250 \times {10^{ - 3}}\,L$.
The molarity of the $NaOH$solution $= \,0.205\,M$.
Now, the equation $\left( 1 \right)$ can be written as,
$No.\,of\,moles\,of\,NaOH\, = \,\,Molarity \times Volume\,of\,solution\,\left( {in\,L} \right)$
Putting the values we get,
$No.\,of\,moles\,of\,NaOH\, = \,\left( {250 \times {{10}^{ - 3}}} \right)\,L \times 0.205\,mol\,{L^{ - 1}}\,\, = \,\,0.05125\,mol$
Therefore the solution contains $0.00285\,moles$of$NaOH$.
We know, $No.\,of\,moles\,of\,a\,compound\, = \,\dfrac{{Given\,Mass}}{{Molar\,Mass}}$.
$\Rightarrow \,Given\,Mass\, = \,No.\,of\,moles\,of\,the\,compound \times \,Molar\,Mass$
Now, the molar mass of $NaOH\, = \,40\,g\,mo{l^{ - 1}}$.
$\therefore$ Mass of $NaOH$needed to prepare $250\,mL$of $0.205\,M\,\,NaOH$
$\, = \,0.05125\,mol \times 40\,g\,mo{l^{ - 1}}\, = \,2.05\,g$
Hence the correct answer is (iii) $2.05\,g$.
Additional Information:
Molarity is a unit to express the concentration of a solution. As mentioned earlier it is given by the ratio of number of moles of solute to the volume of the solution in $L$. Since it is related to the volume of the solution it is temperature $($as volume is dependent upon temperature$)$.

Note: Always remember in expression of molarity the volume of the solution is in$L$. Make sure to check the units properly so that you do not mix up the SI and CGS units. Do the calculation in a stepwise manner in order to avoid errors as you can keep track of where the same units are cancelling each other if done one step at a time.