
How many moles of $ N{a^ + } $ ions are present in $ 275.0\;{\text{mL}} $ of $ 0.35{\text{M}} $ $ N{a_3}P{O_4} $ solution?
Answer
551.7k+ views
Hint: The total number of moles of solvent per litre of solution is defined as the molarity of a given solution. The molality of a solution depends on the changes in the system's physical properties, such as pressure and temperature, as the system volume changes with the change in the system's physical conditions, unlike mass. We shall calculate the moles of the compound present and thus, the moles of sodium present.
Formula Used
We would require the formula for molarity to solve this question
$ {\text{molarity = }}\dfrac{{{n_s}}}{{{v_s}}} $
Where
$ {n_s} $ is the number of moles of solute
$ {v_s} $ is the volume of solvent in litres.
Complete Step-by-Step Solution
According to the question, the following information is provided to us
Molarity of $ N{a_3}P{O_4} $ solution $ = 0.35 {\text{M}} $
Volume of $ N{a_3}P{O_4} $ solution, $ {v_s} = 275 {\text{mL}} = 0.275{\text{L}} $
We know that molarity tells us the number of moles of solute per unit volume of solvent (in litres). Now, we will use the above formula to get the result.
$ {\text{molarity}} = \dfrac{{{n_s}}}{{{v_s}}} $
We can rewrite it as
$ {n_s} = {\text{molarity}} \times {v_s} $
Upon substituting values, we get
$ {n_s} = 0.35{\text{M}} \times 0.275{\text{L}} $
Upon solving, we get $ {n_s} = 9.6 \times {10^{ - 2}} $ moles of $ N{a_3}P{O_4} $
Hence, $ {n_s} = 9.6 \times {10^{ - 2}} $ moles of $ N{a_3}P{O_4} $ are present in $ 0.35{\text{M}} $ $ N{a_3}P{O_4} $ solution.
Additional Information
Molarity, which is referred to as molar, is represented by M. The molarity of a solution where one gram of solute is dissolved in a litre of solution is one molar. As we know, the solvent and solute blend is used in a solution to form a solution, so the total volume of the solution is taken.
Note
In solutions, the majority of reactions occur and it is therefore essential to understand how the amount of substance is expressed when it is present in the solution. The quantity of substances in the solution is expressed in many ways.
Formula Used
We would require the formula for molarity to solve this question
$ {\text{molarity = }}\dfrac{{{n_s}}}{{{v_s}}} $
Where
$ {n_s} $ is the number of moles of solute
$ {v_s} $ is the volume of solvent in litres.
Complete Step-by-Step Solution
According to the question, the following information is provided to us
Molarity of $ N{a_3}P{O_4} $ solution $ = 0.35 {\text{M}} $
Volume of $ N{a_3}P{O_4} $ solution, $ {v_s} = 275 {\text{mL}} = 0.275{\text{L}} $
We know that molarity tells us the number of moles of solute per unit volume of solvent (in litres). Now, we will use the above formula to get the result.
$ {\text{molarity}} = \dfrac{{{n_s}}}{{{v_s}}} $
We can rewrite it as
$ {n_s} = {\text{molarity}} \times {v_s} $
Upon substituting values, we get
$ {n_s} = 0.35{\text{M}} \times 0.275{\text{L}} $
Upon solving, we get $ {n_s} = 9.6 \times {10^{ - 2}} $ moles of $ N{a_3}P{O_4} $
Hence, $ {n_s} = 9.6 \times {10^{ - 2}} $ moles of $ N{a_3}P{O_4} $ are present in $ 0.35{\text{M}} $ $ N{a_3}P{O_4} $ solution.
Additional Information
Molarity, which is referred to as molar, is represented by M. The molarity of a solution where one gram of solute is dissolved in a litre of solution is one molar. As we know, the solvent and solute blend is used in a solution to form a solution, so the total volume of the solution is taken.
Note
In solutions, the majority of reactions occur and it is therefore essential to understand how the amount of substance is expressed when it is present in the solution. The quantity of substances in the solution is expressed in many ways.
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

