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Question: Active mass of 0.64 g \(S{{O}_{2}}\) in 10 lit vessel is A. \({{10}^{-2}}M\) B. \({{10}^{-3}}M\...

Active mass of 0.64 g SO2S{{O}_{2}} in 10 lit vessel is
A. 10−2M{{10}^{-2}}M
B. 10−3M{{10}^{-3}}M
C. 10−1M{{10}^{-1}}M
D. 0.64 g

Explanation

Solution

. To solve this question, we must know the meaning of the active mass of a compound. Active mass of a compound is nothing but the molarity of the given solution. So basically we need to find the molarity of the solution.

Complete step by step answer:
-Molarity is defined as the ratio of the moles of the solute to the volume of the solution and can be expressed mathematically as
molarity=moles of solutevolume of solution=molLmolarity = \dfrac{moles\text{ }of\text{ }solute}{volume\text{ of solution}} = \dfrac{mol}{L}

-Now to find the molarity, we first need to find the moles of the solute. Only then we can find the molarity of the solution. Mole is the unit for measuring the amount of a substance. It is given by the formula
Moles of solute = wt. in gramsmolecular wt.\dfrac{wt.\text{ in grams}}{molecular\text{ wt}\text{.}}

-As the numerator and the denominator both have the same unit, the quantity ‘mole’ is unitless. For the given question, we can see that we are given the weight of SO2S{{O}_{2}} to be 0.64 grams. The molecular weight of SO2S{{O}_{2}} will be the sum of the molecular weights of S and both the oxygen atoms which can be shown as
Molecular weight of SO2S{{O}_{2}} = 32 + 16 + 16 = 64 grams.
So, the moles will be given as
Moles of solute = wt. in gramsmolecular wt.=0.6464=10−2\dfrac{wt.\text{ in grams}}{molecular\text{ wt}\text{.}}=\dfrac{0.64}{64}={{10}^{-2}}

-So, molarity=moles of solutevolume of solution=10−210=10−3molarity = \dfrac{moles\text{ }of\text{ }solute}{volume\text{ of solution}} = \dfrac{{{10}^{-2}}}{10}={{10}^{-3}}
So, the correct answer is “Option B”.

Note: Always take into account the unit of all the quantities present in the question. Remember that 1M= mol/L. No other unit defines 1M. Also, keep into account the relationship between the units of mass and the units of volume. 1L = 1000ml and 1ml = 1cc. Also, volume has to be taken of the solution and the moles are taken of the solute.