
What is the weight by volume percentage of \[3N\text{ }{{H}_{2}}S{{O}_{4}}\] solution?
A.$5%$
B.$10%$
C.$14.7%$
D.$20%$
Answer
518.1k+ views
Hint: We know that the concentration of a solution can be measured with the help of normality. Normality is the ratio of a number of grams equivalent by volume of solution in litre. It can measure the number of ions in the precipitation reaction.
Complete answer:
In the mole concept we have studied about the molarity and we have studied that molar concentration of substance is also known as molarity. It is defined as the ratio of moles of substance to volume of solution in litre. Mole is defined as weight in grams divided by the molecular weight of the substance. So we will find out the molarity by the formula; M = Number of moles of solute /volume of the solution in litre
Normality is used for the acid and bases mostly. The normality is also related to the molarity and normality is equal to the product of n and molarity where n is equal to the number of hydrogen ions in the acid or no. of hydroxyl ions in the base or n-factor. Once the number of moles is determined we calculate molarity of sulphuric acid easily with the help of molarity formula.
\[Normality\text{ }=\text{ }\dfrac{Weight}{Equivalent\text{ }Weight}\times \dfrac{1}{V(L)}\]
Substituting the given values;
\[3N\text{ }=\text{ }\dfrac{x}{\left( \dfrac{98}{2} \right)}\times \dfrac{1000}{V(mL)}\] Here it is multiplied by $1000$ since we have done from L to mL.
On further solving we get;
\[\dfrac{x}{V}\times 1000=\dfrac{3\times 98}{2}\]
From the above equation, we get;
\[\Rightarrow \dfrac{x}{V}\times 100=\dfrac{147}{10}\]
Therefore, \[\Rightarrow \dfrac{w}{V}%=14.7%\]
Therefore, the correct answer is option C.
Note:
Remember that in mole problems we should always remember some important relation .Here in this problem we have used some relations to solve the problem as we have found the mass with the help of given density and then put the value of mass in the formula to find the number of moles.
Complete answer:
In the mole concept we have studied about the molarity and we have studied that molar concentration of substance is also known as molarity. It is defined as the ratio of moles of substance to volume of solution in litre. Mole is defined as weight in grams divided by the molecular weight of the substance. So we will find out the molarity by the formula; M = Number of moles of solute /volume of the solution in litre
Normality is used for the acid and bases mostly. The normality is also related to the molarity and normality is equal to the product of n and molarity where n is equal to the number of hydrogen ions in the acid or no. of hydroxyl ions in the base or n-factor. Once the number of moles is determined we calculate molarity of sulphuric acid easily with the help of molarity formula.
\[Normality\text{ }=\text{ }\dfrac{Weight}{Equivalent\text{ }Weight}\times \dfrac{1}{V(L)}\]
Substituting the given values;
\[3N\text{ }=\text{ }\dfrac{x}{\left( \dfrac{98}{2} \right)}\times \dfrac{1000}{V(mL)}\] Here it is multiplied by $1000$ since we have done from L to mL.
On further solving we get;
\[\dfrac{x}{V}\times 1000=\dfrac{3\times 98}{2}\]
From the above equation, we get;
\[\Rightarrow \dfrac{x}{V}\times 100=\dfrac{147}{10}\]
Therefore, \[\Rightarrow \dfrac{w}{V}%=14.7%\]
Therefore, the correct answer is option C.
Note:
Remember that in mole problems we should always remember some important relation .Here in this problem we have used some relations to solve the problem as we have found the mass with the help of given density and then put the value of mass in the formula to find the number of moles.
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

