
If \[4{\text{g}}\] of \[{\text{NaOH}}\] are contained in one deciliter of solution, its molarity would be:
A.\[4{\text{M}}\]
B.\[2{\text{M}}\]
C.\[1{\text{M}}\]
D.\[1.5{\text{M}}\]
Answer
548.7k+ views
Hint: To find molarity, we need to calculate moles of solute present in per litre of the solution. Moles of a substance is a SI unit of amount of substance, it can be calculated from mass, volume and number of a substance.
Complete step by step answer:
In order to find molarity of \[{\text{NaOH}}\] or sodium hydroxide, we need to first find moles of sodium hydroxide present in the given volume of solution. As the given mass of sodium hydroxide is \[4{\text{g}}\] and as molar mass of sodium hydroxide means mass of 1 mole sodium hydroxide molecules and that is \[40{\text{g}}\] , from this data we can easily calculate moles of sodium hydroxide and moles of a substance can be given as:
\[{\text{no of moles}} = \dfrac{{{\text{given mass}}}}{{{\text{molar mass}}}}\] ,
putting values of mass of sodium hydroxide given and molar mass of sodium hydroxide.
We get, number of moles of sodium hydroxide as: \[{\text{no of moles}} = \dfrac{4}{{40}} = 0.1{\text{mol}}\] .
Now volume of the solution can be determine from the given data in question as it is given that \[{\text{NaOH}}\] are contained in one deciliter of solution, it means volume of solution is \[0.1{\text{L}}\] .
Now molarity of the solution is defined as the number of moles of a solute present in \[1{\text{L}}\] or \[1000{\text{mL}}\] of the solution. In our case, solute is sodium hydroxide, thereby molarity of the solution can be given as:
\[{\text{Molarity}} = \dfrac{{{\text{moles of NaOH}}}}{{{\text{Volume of solution}}\left( {{\text{in L}}} \right)}} = \dfrac{{0.1}}{{0.1}} = 1{\text{M}}\] .
Thus, the correct option is C.
Note:
Molality of a solution is defined as the number of moles of solute present in \[1000{\text{g}}\] or \[1{\text{kg}}\] of a solvent. Mole fraction is the ratio of number of moles of solute or solvent present in the solution and the total number of moles present in the solution of the substance.
Complete step by step answer:
In order to find molarity of \[{\text{NaOH}}\] or sodium hydroxide, we need to first find moles of sodium hydroxide present in the given volume of solution. As the given mass of sodium hydroxide is \[4{\text{g}}\] and as molar mass of sodium hydroxide means mass of 1 mole sodium hydroxide molecules and that is \[40{\text{g}}\] , from this data we can easily calculate moles of sodium hydroxide and moles of a substance can be given as:
\[{\text{no of moles}} = \dfrac{{{\text{given mass}}}}{{{\text{molar mass}}}}\] ,
putting values of mass of sodium hydroxide given and molar mass of sodium hydroxide.
We get, number of moles of sodium hydroxide as: \[{\text{no of moles}} = \dfrac{4}{{40}} = 0.1{\text{mol}}\] .
Now volume of the solution can be determine from the given data in question as it is given that \[{\text{NaOH}}\] are contained in one deciliter of solution, it means volume of solution is \[0.1{\text{L}}\] .
Now molarity of the solution is defined as the number of moles of a solute present in \[1{\text{L}}\] or \[1000{\text{mL}}\] of the solution. In our case, solute is sodium hydroxide, thereby molarity of the solution can be given as:
\[{\text{Molarity}} = \dfrac{{{\text{moles of NaOH}}}}{{{\text{Volume of solution}}\left( {{\text{in L}}} \right)}} = \dfrac{{0.1}}{{0.1}} = 1{\text{M}}\] .
Thus, the correct option is C.
Note:
Molality of a solution is defined as the number of moles of solute present in \[1000{\text{g}}\] or \[1{\text{kg}}\] of a solvent. Mole fraction is the ratio of number of moles of solute or solvent present in the solution and the total number of moles present in the solution of the substance.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

Which animal has three hearts class 11 biology CBSE

10 examples of friction in our daily life

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

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

