Question

How many moles of $NaOH$are present in $19.0\,mL$ of $0.150\,M\,NaOH\,?$
A) $0.00355\,moles$
B) $0.00195\,moles$
C) $0.00285\,moles$
D) $0.00485\,moles$

Answer 1

The molarity or molar concentration of a solute is defined as the number of moles of solute per liter of solution and is denoted by $M$ and 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. Convert the units as required and find the number of moles of the solute i.e. $NaOH$.

Complete 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 $= \,19.0\,mL$
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$NaOH\, = \,19.0\,mL \times \dfrac{{1\,L}}{{1000\,mL}}\, = \,19 \times {10^{ - 3}}\,L$.
The molarity of the $NaOH$solution $= \,0.150\,M$.
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,
$No.\,of\,moles\,of\,NaOH\, = \,\left( {19.0 \times {{10}^{ - 3}}} \right)\,L \times 0.150\,mol\,{L^{ - 1}}\,\, = \,\,0.00285\,mol$
Therefore the solution contains $0.00285\,moles$of$NaOH$.

Hence the correct answer is option (C) $0.00285\,moles$.

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.