Question

The molarity of a solution containing $1g$$NaOH$in $250ml$ Of solution is
A.$0.1M$ [EAMCET$1990$]
B.$1M$
C.$0.01M$
D.$0.001M$

Answer 1

Hint: The molarity of any solution is expressed by the number of moles present in one liter of solution. To approach this question first we need to calculate the number of moles present in $NaOH$the solution. The number of moles is the ratio of the given mass to the molar mass of the component.

Formula Used:Number of moles of any component, $n=\dfrac{m}{M}$
Where $m$ and $M$ denote the given mass and molar mass of the component.
Molarity,$=\dfrac{n}{V(Litre)}$
Here $n=$number of moles of any component.
$V=$The volume of the solution in $litre$

Complete answer:Here the solution contains $1g$$NaOH$ and number of moles of sodium hydroxide$NaOH$can be calculated as follows,
The molar mass of sodium hydroxide,${{M}_{NaOH}}=$(Atomic weight of $Na$)$+$(Atomic weight of oxygen)$+$(Atomic weight of Hydrogen)
$\therefore {{M}_{NaOH}}=(40+16+1)=57g/mol$
Given the mass of sodium hydroxide,${{m}_{NaOH}}$ $=1g$
Number of moles of $NaOH$,${{n}_{NaOH}}=\dfrac{{{m}_{NaOH}}}{{{M}_{NaOH}}}=\dfrac{1}{40}=0.025mol$
The volume of the solution,$V=250ml$
As we know $1L=1000ml$ , hence $V=\dfrac{250}{1000}=0.25L$
Therefore molarity,$=\dfrac{{{n}_{NaOH}}}{V}=\dfrac{0.025mol}{0.25L}=0.1mol/L$

Thus, option (A) is correct.

Additional information: Molarity and Molality are two different terms. Molarity of any solution is the number of moles of any component present per liter of the solution but Molality is the number of moles present per kg of solvent. Molarity is temperature dependent but molality does not depend on temperature. There in the formula of molarity, the term volume is included and temperature affects volume. But in molality only the mass term is present and the mass of any substance does not change with a temperature change.

Note: To approach this type of problem we must know about the mole concept. Because this is a fundamental concept in whole chemistry. It is very difficult sometimes when we deal with microparticles whose sizes are very small therefore defining a mole helps us to convert moles into their equivalent particles or grams to their equivalent moles.