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Active mass of a 6 % solution of compound X is 2 its molecular weight will be

Last updated date: 18th Jun 2024
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Hint: The molarity of the solution is commonly regarded as the active mass of the substance. Molarity is the number of moles of the solute present in 1 litre of the solution. The unit of the molarity is mol/litre. It is usually indicated within the square brackets,[].

Complete step by step answer:
The law of mass action is used in such cases which states that-The rate of the chemical reaction is directly proportional to the masses of the reacting substances (reactants) with each product's mass raised to the power of the stoichiometric coefficient of the respective reactant in the chemical equation.
Suppose for a hypothetical reaction the rate, the rate of the reaction is calculated as:
  aA + bB \to xX + yY \\
  {k_C} = \dfrac{{{{[X]}^x}{{[Y]}^y}}}{{{{[A]}^a}{{[B]}^b}}} \\
\end{gathered} $
The active mass of the substance is calculated as {ActiveMass = $\dfrac{{\text{Number Of Moles}}}{{\text{volume of solution in litres}}}$

And the term number of moles is calculated as Number Of Moles = $\dfrac{{\text{Weight Of Solute}}}{{\text{Molecular Weight}}}$

As per the question 6% solution of compound X is given which means 6 g of solute is dissolved in 100 ml of the solution.
Thus mass of the solute present in 1 litre of solution=60 g
$Concentration = \dfrac{{Moles}}{\begin{gathered}
  Litre \\
   = \dfrac{{60}}{{Molecular Weight}} \\
\end{gathered} }$
Given active mass =2
Thus Molecular weight = Number of moles/concentration
Molecular weight = 30
Thus the molecular weight of the compound X will be 30

So, the correct answer is Option A.

Two aspects were initially involved in the formulation of the formula of the law of mass action that is the kinetics and the equilibrium aspects of the reaction were involved in the law of mass action formulation.