
A stock solution of $10M,NaOH$ was used to prepare $2litre$ of $0.5M,NaOH$. How many milliliters of $NaOH$ stock solution will be needed?
A.$10mL$
B.$100mL$
C.$1000mL$
D.$200mL$
E.$2000mL$
Answer
579.3k+ views
Hint:
A stock solution is a concentrated solution which can be diluted with any other solvent as per the requirement. Dilution decreases the concentration. Molarity is a way of expressing concentration of the solution. Molarity expresses the concentration of solution in terms of the number of moles present in a given volume of solution.
Complete step by step answer:
Generally in laboratories stock solutions of various compounds are prepared. These stock solutions are concentrated solutions of that compound and this stock solution can be diluted as per the requirement with an appropriate solvent. After dilution their concentration gets lowered.
We know that there are different ways of expressing concentration. Molarity, Normality, Molality are some of them. In simple words molarity can be defined as the number of moles of solute present or dissolved in one litre of solution. Molarity is denoted by $M$ . The formula for molarity can be given as-
$Molarity = \dfrac{\text{number of moles of solute}}{\text{Volume of solution in litres}}$
For the solutions of the same compound, the relation between their molarity and volume can be given as-
${M_1}{V_1} = {M_2}{V_2}$
Where ${M_1}$ and ${V_1}$ are the molarity and volume of solution $1$ and ${M_2}$ and ${V_2}$ are the molarity and volume of solution $2$ .
Let the $2litre$ of $0.5M,NaOH$ solution be solution $1$ . Then,
${M_1} = 0.5M$
${V_1} = 2litres = 2000mL$
Let the stock solution of $10M,NaOH$ be a solution $2$ . Then,
${M_2} = 10M$
We need to calculate ${V_2}$ .
Now substituting these values in the above equation-
${M_1}{V_1} = {M_2}{V_2}$
$0.5 \times 2000 = 10 \times {V_2}$
${V_2} = \dfrac{{0.5 \times 2000}}{{10}}$
${V_2} = 100mL$
So $100mL$ of stock solution of $10M,NaOH$ will be required to prepare $2litre$ of $0.5M,NaOH$ solution.
So the correct option is B.
Note:Molarity depends on temperature. This is because as we know the volume of a solution depends on temperature. Molality is defined as the number of moles of solute that are dissolved or present in $1kg$ of solvent. Molality is not dependent on temperature as mass does not depend on temperature.
A stock solution is a concentrated solution which can be diluted with any other solvent as per the requirement. Dilution decreases the concentration. Molarity is a way of expressing concentration of the solution. Molarity expresses the concentration of solution in terms of the number of moles present in a given volume of solution.
Complete step by step answer:
Generally in laboratories stock solutions of various compounds are prepared. These stock solutions are concentrated solutions of that compound and this stock solution can be diluted as per the requirement with an appropriate solvent. After dilution their concentration gets lowered.
We know that there are different ways of expressing concentration. Molarity, Normality, Molality are some of them. In simple words molarity can be defined as the number of moles of solute present or dissolved in one litre of solution. Molarity is denoted by $M$ . The formula for molarity can be given as-
$Molarity = \dfrac{\text{number of moles of solute}}{\text{Volume of solution in litres}}$
For the solutions of the same compound, the relation between their molarity and volume can be given as-
${M_1}{V_1} = {M_2}{V_2}$
Where ${M_1}$ and ${V_1}$ are the molarity and volume of solution $1$ and ${M_2}$ and ${V_2}$ are the molarity and volume of solution $2$ .
Let the $2litre$ of $0.5M,NaOH$ solution be solution $1$ . Then,
${M_1} = 0.5M$
${V_1} = 2litres = 2000mL$
Let the stock solution of $10M,NaOH$ be a solution $2$ . Then,
${M_2} = 10M$
We need to calculate ${V_2}$ .
Now substituting these values in the above equation-
${M_1}{V_1} = {M_2}{V_2}$
$0.5 \times 2000 = 10 \times {V_2}$
${V_2} = \dfrac{{0.5 \times 2000}}{{10}}$
${V_2} = 100mL$
So $100mL$ of stock solution of $10M,NaOH$ will be required to prepare $2litre$ of $0.5M,NaOH$ solution.
So the correct option is B.
Note:Molarity depends on temperature. This is because as we know the volume of a solution depends on temperature. Molality is defined as the number of moles of solute that are dissolved or present in $1kg$ of solvent. Molality is not dependent on temperature as mass does not depend on temperature.
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