What volume of water is to be added to $100c{m^3}$ of $0.5M$ $NaOH$ solution to make it $0.1M$ solution?
A. $200c{m^3}$
B. $400c{m^3}$
C. $500c{m^3}$
D. $100c{m^3}$
Answer
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Hint: We have to know that dilution is the way toward diminishing the convergence of a solute in an answer, ordinarily basically by blending in with something more dissolvable like adding more water to the arrangement. To weaken an answer intends to add more dissolvable without the option of more solute. The subsequent arrangement is completely blended in order to guarantee that all pieces of the arrangement are indistinguishable.
Complete answer:
We have to know that a similar direct relationship applies to gases and fumes weakened in air for instance. Although thorough blending of gases and fumes may not be as effectively accomplished. The ${M_1}{V_1}$ rises to ${M_2}{V_2}$ condition is utilized to figure weakening. ${M_1}$ is the molarity and ${V_1}$ is the volume of the concentrated arrangement. ${M_2}$ is the molarity and ${V_{}}$ is the volume of the weakened arrangement.
Numerically this relationship can be displayed by condition.
${M_1} \times {V_1} = {M_2}{V_2}$
Where, in the given details,
${M_1}$ = The initial concentration of molarity = $0.5M$
${M_2}$ = The final concentration of molarity = $0.1M$
${V_1}$ = The initial volume = $100c{m^3}$
${V_2}$ = The final volume
Using the above expression, calculate the final volume of the solution.
${M_1} \times {V_1} = {M_2}{V_2}$
Applying given values in the above equation,
$0.5 \times 100 = 0.1 \times {V_2}$
Then,
${V_2} = 500c{m^3}$
Therefore,
$V = 500 - 100 = 400c{m^3}$
Hence, the volume of the water is $400c{m^3}$ .
So, the correct answer is “Option B”.
Note:
We have to know that, ${M_1} \times {V_1} = {M_2}{V_2}$ , which joins both the mole idea and the convergence of arrangements, is regularly utilized in science computations. Factors, for example, reasonable agreement and critical thinking abilities add to understudies presentation in ascertaining issues identified with the centralization of solutions rated arrangement.
Complete answer:
We have to know that a similar direct relationship applies to gases and fumes weakened in air for instance. Although thorough blending of gases and fumes may not be as effectively accomplished. The ${M_1}{V_1}$ rises to ${M_2}{V_2}$ condition is utilized to figure weakening. ${M_1}$ is the molarity and ${V_1}$ is the volume of the concentrated arrangement. ${M_2}$ is the molarity and ${V_{}}$ is the volume of the weakened arrangement.
Numerically this relationship can be displayed by condition.
${M_1} \times {V_1} = {M_2}{V_2}$
Where, in the given details,
${M_1}$ = The initial concentration of molarity = $0.5M$
${M_2}$ = The final concentration of molarity = $0.1M$
${V_1}$ = The initial volume = $100c{m^3}$
${V_2}$ = The final volume
Using the above expression, calculate the final volume of the solution.
${M_1} \times {V_1} = {M_2}{V_2}$
Applying given values in the above equation,
$0.5 \times 100 = 0.1 \times {V_2}$
Then,
${V_2} = 500c{m^3}$
Therefore,
$V = 500 - 100 = 400c{m^3}$
Hence, the volume of the water is $400c{m^3}$ .
So, the correct answer is “Option B”.
Note:
We have to know that, ${M_1} \times {V_1} = {M_2}{V_2}$ , which joins both the mole idea and the convergence of arrangements, is regularly utilized in science computations. Factors, for example, reasonable agreement and critical thinking abilities add to understudies presentation in ascertaining issues identified with the centralization of solutions rated arrangement.
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