Question

20 ml of 0.1 M, 30 ml of 0.2 M and 30 ml of 0.3 M
solutions of oxalic acid are mixed and the volume
is made 100 ml. The molarity of the resulting
solution is
(2) 8.51 M
(1) 0.21 M
(3) 5.67 M
(4) 0.17 M​

Answer 1

Answer:

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Answer 2

$\rm \huge {Answer:-}$

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$\sf \red{\underline{\underline{Given:-}}}$

★20 ml of 0.1 M, 30 ml of 0.2 M and 30 ml of 0.3 M solutions of oxalic acid are mixed and the volume is made 100 ml.

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$\sf \blue{\underline{\underline{To\: Find:-}}}$

★The molarity of the resulting solution=?

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$\sf \pink{\underline{\underline{Consideration:-}}}$

★Now,let's say 3 solutions of oxalic acid with certain volumes are mixed and the total volume is made up to 100ml.

★Solution 1+Solution 2+solution 3=Final solution

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$\sf \purple{\underline{\underline{We\: know:-}}}$

$\small \implies{Molarity(M)=\dfrac{No\: of\: moles(n)}{Volume\: in\: litres(v)}}$

★As the molarity and volume of the solutions are given,no of moles are to be determined.

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★Solution 1:-

$\dashrightarrow{20\: ml\:of\: 0.1\: M}$

$\bf {Molarity (M_{1})=0.1}$

$\bf {volume(v_{1})=20ml}$

★As we need the volume in litres,

$\bf {1ml=\frac{1}{1000}litres}$

$\bf {v_{1}=\frac{20}{1000}litres}$

$\bf {v_{1}=0.02litres}$

$\colon \dashrightarrow\bf{Molarity(M_{1})=\frac{n_{1}}{v_{1}}}$

$\colon \dashrightarrow\bf{0.1=\frac{n_{1}}{0.02litres}}$

$\colon \dashrightarrow\bf{n_{1}=0.1\times0.02litres}$

$\implies \bf\green{n_{1}=0.002}$

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★Solution 2:-

$\dashrightarrow{30\: ml\:of\: 0.2\: M}$

$\bf {Molarity (M_{2})=0.2}$

$\bf {volume(v_{2})=30ml}$

$\bf {v_{2}=\frac{30}{1000}litres}$

$\bf {v_{1}=0.03litres}$

$\colon \dashrightarrow\bf{Molarity(M_{2})=\frac{n_{2}}{v_{2}}}$

$\colon \dashrightarrow\bf{0.2=\frac{n_{2}}{0.03litres}}$

$\colon \dashrightarrow\bf{n_{2}=0.2\times0.03litres}$

$\implies \bf \green{n_{2}=0.006}$

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★Solution 3:-

$\dashrightarrow\bf{30\: ml\:of\: 0.3\: M}$

$\bf {Molarity (M_{3})=0.3}$

$\bf{volume(v_{3})=30ml}$

$\bf {v_{3}=\frac{30}{1000}litres}$

$\bf {v_{3}=0.03litres}$

$\colon \dashrightarrow\bf{Molarity(M_{3})=\frac{n_{3}}{v_{3}}}$

$\colon \dashrightarrow\bf{0.3=\frac{n_{3}}{0.03litres}}$

$\colon \dashrightarrow\bf{n_{3}=0.3\times0.03litres}$

$\implies \bf \green{n_{3}=0.009}$

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★Now the total no of moles,

$\dashrightarrow \bf{n_{total}=n_{1}+n_{2}+n_{3}}$

$\dashrightarrow\bf {n_{total}=0.002+0.006+0.009}$

$\dashrightarrow\bf {n_{total}=0.017}$

$\dashrightarrow {Total\: volume\: given:-100ml}$

$\bf {v_{total}=\frac{100}{1000}litres}$

$\bf {v_{total}=0.1litres}$

★The total molarity comes out to be,

$\colon \dashrightarrow\bf{Molarity(M_{total})=\frac{n_{total}}{v_{total}}}$

$\colon \dashrightarrow\bf{Molarity(M_{total})=\frac{0.017}{0.1litres}}$

$\colon \dashrightarrow\bf{Molarity(M_{total})=\frac{17\times10}{1000}}$

$\implies \bf\green {\fbox{Total\: molarity=0.17M}}$

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$\sf \orange{\underline{\underline{Henceforth,}}}$

★Option (4)-0.17M is the correct answer