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The number of gram of anhydrous \[N{{a}_{2}}C{{O}_{3}}\] present in 250 ml of 0.25 M solution is :
(A) 6.625 g
(B) 0.625 g
(C) 0.567 g
(D) 7.125 g

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Last updated date: 20th Apr 2024
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Answer
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Hint: ‘Molarity’ denoted as M or molar concentration is the number of moles of a solute present per litre of the solution.
Molar mass of anhydrous\[N{{a}_{2}}C{{O}_{3}}\]= \[\text{= }\!\![\!\!\text{ (2 }\times 23)\text{ + (12) + (3 }\times \text{ 16) }\!\!]\!\!\text{ g/mol }\]
\[\text{= 106 g/mol }\]
Using this value and the definition of molarity, find out the amount of sample present in the solution.
Formula used:
As we know, the formula of molarity is:
\[\begin{align}
& molarity(M)\text{ = }\dfrac{mass\text{ }of\text{ }solute\text{ }present\text{ }in\text{ }the\text{ }solution\text{ }(g)}{molecular\text{ }mass\text{ }of\text{ }the\text{ }solute\text{ }(g/mol)\text{ }\times \text{ }volume\text{ }of\text{ }the\text{ }solution\text{ }(L)} \\
& \text{ = }\dfrac{number\text{ }of\text{ }moles\text{ }of\text{ }solute\text{ }present\text{ }in\text{ }the\text{ }solution\text{ }(mol)}{\text{ }volume\text{ }of\text{ }the\text{ }solution\text{ }(L)} \\
\end{align}\]

Complete step by step answer:
Given, volume of the solution = 250 ml = 0.25 L
Molarity of the solution (M) = 0.25 M
From the formula, molar mass of the solute = Molar mass of anhydrous\[N{{a}_{2}}C{{O}_{3}}\]
= 106 g/mol
Let, mass of anhydrous\[N{{a}_{2}}C{{O}_{3}}\]present in the solution = x
So, using the formula we get,
\[\text{0}\text{.25 = }\dfrac{x}{106\text{ }\times \text{ 0}\text{.25}}\]
\[\Rightarrow \text{ x = 0}\text{.25 }\times 0.25\text{ }\times \text{ 106}\]
\[\Rightarrow \text{ x = 6}\text{.625}\]
So, the mass of anhydrous\[N{{a}_{2}}C{{O}_{3}}\]present in the solution is 6.625 g.
So, the correct option is A.
Additional information: Sodium carbonate- disodium salt of carbonic acid. It exists in three forms of hydrates:
\[N{{a}_{2}}C{{O}_{3}}.\text{10}{{H}_{2}}O\](sodium carbonate decahydrate)
\[N{{a}_{2}}C{{O}_{3}}.7{{H}_{2}}O\](sodium carbonate heptahydrate)
\[N{{a}_{2}}C{{O}_{3}}.{{H}_{2}}O\](sodium carbonate monohydrate)
The given compound in the question is the anhydrous salt (salt without any water molecule). Another name for this anhydrous salt is calcined soda. It is manufactured in the last step of the Solvay process when sodium hydrogen carbonate is heated in absence of air.

Note: Molar mass is important in analysing the results of experiments. One should be careful about the units. It should be in one system of units.
In an experiment, if two equal amounts of moles of different substances have different volumes, then it implies that the substance with the more volume is larger than the substance with the smaller volume.