
Given,$10g$ of $NaOH$ is mixed with $36g$ of ${H_2}O$ . What is the mole fraction of $NaOH$ in solution?
A. $0.125$
B. $0.111$
C. $0.72$
D. $0.5$
Answer
504.9k+ views
Hint: To figure the quantity of moles of any substance present in the example, we partition the given load of the substance by its molar mass. The mole portion is the quantity of moles of a specific part in the arrangement isolated by the absolute number of moles in the given arrangement.
By using the following formula,
$No.{\text{ofmoles = }}\dfrac{{Mass}}{{Molar{\text{mass}}}}$
$Mole{\text{ fraction = }}\dfrac{{No.{\text{ of moles of NaOH}}}}{{No.{\text{ of moles of NaOH + No}}{\text{. of moles of }}{{\text{H}}_2}O}}$
Complete answer:
The given details are,
The mass of sodium hydroxide = $10g$
The mass of water = $36g$
We have to calculate the mole fraction of sodium hydroxide in solution.
Using the following expression,
First, we have to calculate the number of moles of sodium hydroxide, and water.
For sodium hydroxide,
$No.{\text{ of moles = }}\dfrac{{Mass}}{{Molar{\text{ mass}}}} = \dfrac{{10g}}{{40g}}$
Where,
The molar mass of sodium hydroxide = $40g$
Therefore,
The number of moles of sodium hydroxide = $0.25mol$ .
For water,
$No.{\text{ of moles = }}\dfrac{{Mass}}{{Molar{\text{ mass}}}} = \dfrac{{36g}}{{18g}}$
Where,
The molar mass of water = $18g$
Therefore,
The number of moles of water = $2mol$ .
Then, we have using the mole fraction formula,
$Mole{\text{ fraction = }}\dfrac{{No.{\text{ of moles of NaOH}}}}{{No.{\text{ of moles of NaOH + No}}{\text{. of moles of }}{{\text{H}}_2}O}}$
Applying all the values in the above expression,
$Mole{\text{ fraction = }}\dfrac{{0.25}}{{0.25 + 2}}$
On simplification we get,
Mole fraction = $0.111$
Therefore, the mole fraction of sodium hydroxide is $0.111$ .
So, the correct answer is “Option B”.
Note:
We have to know that the mole fraction isn't reliant upon the temperature. The data about the thickness of the stage isn't important to ascertain the mole portion. On account of an optimal gas blend, the mole division is addressed by the proportion of fractional strain to add up to the pressing factor of the combination. Mole part is supposed to be a unit-less and dimensionless articulation.
By using the following formula,
$No.{\text{ofmoles = }}\dfrac{{Mass}}{{Molar{\text{mass}}}}$
$Mole{\text{ fraction = }}\dfrac{{No.{\text{ of moles of NaOH}}}}{{No.{\text{ of moles of NaOH + No}}{\text{. of moles of }}{{\text{H}}_2}O}}$
Complete answer:
The given details are,
The mass of sodium hydroxide = $10g$
The mass of water = $36g$
We have to calculate the mole fraction of sodium hydroxide in solution.
Using the following expression,
First, we have to calculate the number of moles of sodium hydroxide, and water.
For sodium hydroxide,
$No.{\text{ of moles = }}\dfrac{{Mass}}{{Molar{\text{ mass}}}} = \dfrac{{10g}}{{40g}}$
Where,
The molar mass of sodium hydroxide = $40g$
Therefore,
The number of moles of sodium hydroxide = $0.25mol$ .
For water,
$No.{\text{ of moles = }}\dfrac{{Mass}}{{Molar{\text{ mass}}}} = \dfrac{{36g}}{{18g}}$
Where,
The molar mass of water = $18g$
Therefore,
The number of moles of water = $2mol$ .
Then, we have using the mole fraction formula,
$Mole{\text{ fraction = }}\dfrac{{No.{\text{ of moles of NaOH}}}}{{No.{\text{ of moles of NaOH + No}}{\text{. of moles of }}{{\text{H}}_2}O}}$
Applying all the values in the above expression,
$Mole{\text{ fraction = }}\dfrac{{0.25}}{{0.25 + 2}}$
On simplification we get,
Mole fraction = $0.111$
Therefore, the mole fraction of sodium hydroxide is $0.111$ .
So, the correct answer is “Option B”.
Note:
We have to know that the mole fraction isn't reliant upon the temperature. The data about the thickness of the stage isn't important to ascertain the mole portion. On account of an optimal gas blend, the mole division is addressed by the proportion of fractional strain to add up to the pressing factor of the combination. Mole part is supposed to be a unit-less and dimensionless articulation.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

