
The concentration of an aqueous solution of $0.01M$$C{{H}_{3}}OH$solution is very nearly equal to which of the following [BITS$1992$]
A.$0.01%$$C{{H}_{3}}OH$
B.$0.01m$ $C{{H}_{3}}OH$
C.${{x}_{C{{H}_{3}}OH}}=0.01$
D.$0.99M$ ${{H}_{2}}O$
E.$0.01N$ $C{{H}_{3}}OH$
Answer
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Hint: The concentration of $0.01M$$C{{H}_{3}}OH$means when $0.01$$mol$methanol that is$C{{H}_{3}}OH$is dissolved in $1$$Litre$of aqueous solution. Here in this problem, we have to check the given option which is very nearly equal to the given concentration of the $HCl$solution. For this, we need to understand the term percentage$(%)$, mole fraction $(x)$, molarity ($M$), and normality ($N$).
Complete answer:In this question, the concentration $HCl$is given $0.01M$. As we know when $1mol$ a compound is dissolved in a total volume of the solution $1$$litre$ is called Molarity($M$).In the same way here $0.01mol$of $HCl$is dissolved in a total volume of $1$$litre$ the aqueous solution.
In option (A), $0.01%$$C{{H}_{3}}OH$ that means $0.01g$of $C{{H}_{3}}OH$is dissolved in $100ml$of aqueous solution hence this is not very nearly equal to$0.01M$$C{{H}_{3}}OH$.
In (B) is also not the correct one.
In (C), the mole fraction of $C{{H}_{3}}OH$,${{X}_{C{{H}_{3}}OH}}=0.01$ that indicates the ratio of the number of moles of $HCl$to.
the total sum of the number of moles of ${{H}_{2}}O$and $HCl$. Hence this is not the correct option.
Also (D) is the wrong option.
Finally, in (E),$0.01N$$C{{H}_{3}}OH$the means $0.01gm-equivalent$ $C{{H}_{3}}OH$is dissolved in a total volume of $1$$litre$solution. Also, normality ($N$) can be expressed as
Normality($N$)$=\dfrac{Molarity(M)}{n-factor}$
$n-$the factor is the total number of positive or negative charge present.
$C{{H}_{3}}OH\rightleftharpoons C{{H}_{3}}{{O}^{-}}+{{H}^{+}}$
Total number of $C{{H}_{3}}{{O}^{-}}$$=$ the total number of ${{H}^{+}}$$=1$
Hence $n-$factor of $C{{H}_{3}}OH$is $1$.
Molarity($M$)$=$Normality($N$)$\times $$n-factor=0.01N\times 1=0.01M$
Therefore, $0.01N$$C{{H}_{3}}OH$ is very nearly equal to $0.01M$ $C{{H}_{3}}OH$.
Thus option (E) is correct.
Note: To solve these types of problems we must understand the basic idea about molarity, normality, molality, mole fraction, and percentage. It will help us to approach many problems in chemistry like redox reactions, kinetics, thermodynamics, etc.
Complete answer:In this question, the concentration $HCl$is given $0.01M$. As we know when $1mol$ a compound is dissolved in a total volume of the solution $1$$litre$ is called Molarity($M$).In the same way here $0.01mol$of $HCl$is dissolved in a total volume of $1$$litre$ the aqueous solution.
In option (A), $0.01%$$C{{H}_{3}}OH$ that means $0.01g$of $C{{H}_{3}}OH$is dissolved in $100ml$of aqueous solution hence this is not very nearly equal to$0.01M$$C{{H}_{3}}OH$.
In (B) is also not the correct one.
In (C), the mole fraction of $C{{H}_{3}}OH$,${{X}_{C{{H}_{3}}OH}}=0.01$ that indicates the ratio of the number of moles of $HCl$to.
the total sum of the number of moles of ${{H}_{2}}O$and $HCl$. Hence this is not the correct option.
Also (D) is the wrong option.
Finally, in (E),$0.01N$$C{{H}_{3}}OH$the means $0.01gm-equivalent$ $C{{H}_{3}}OH$is dissolved in a total volume of $1$$litre$solution. Also, normality ($N$) can be expressed as
Normality($N$)$=\dfrac{Molarity(M)}{n-factor}$
$n-$the factor is the total number of positive or negative charge present.
$C{{H}_{3}}OH\rightleftharpoons C{{H}_{3}}{{O}^{-}}+{{H}^{+}}$
Total number of $C{{H}_{3}}{{O}^{-}}$$=$ the total number of ${{H}^{+}}$$=1$
Hence $n-$factor of $C{{H}_{3}}OH$is $1$.
Molarity($M$)$=$Normality($N$)$\times $$n-factor=0.01N\times 1=0.01M$
Therefore, $0.01N$$C{{H}_{3}}OH$ is very nearly equal to $0.01M$ $C{{H}_{3}}OH$.
Thus option (E) is correct.
Note: To solve these types of problems we must understand the basic idea about molarity, normality, molality, mole fraction, and percentage. It will help us to approach many problems in chemistry like redox reactions, kinetics, thermodynamics, etc.
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