The solutions A and B are 0.1 M and 0.2 M i | vedclass

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(C) The number of moles of substance in solution $A$ is $n_A = M_A 	imes V_A = 0.1 \ M 	imes 100 \ mL = 10 \ mmol$. The number of moles of substance

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$0.12$

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(C) The number of moles of substance in solution $A$ is $n_A = M_A 	imes V_A = 0.1 \ M 	imes 100 \ mL = 10 \ mmol$. The number of moles of substance in solution $B$ is $n_B = M_B 	imes V_B = 0.2 \ M 	imes 25 \ mL = 5 \ mmol$. The total number of moles is $n_{total} = 10 \ mmol + 5 \ mmol = 15 \ mmol$. The total volume of the mixture is $V_{total} = 100 \ mL + 25 \ mL = 125 \ mL$. The final molarity $M_{final} = \frac{n_{total}}{V_{total}} = \frac{15 \ mmol}{125 \ mL} = 0.12 \ M$.

List-$I$ (Example)	List-$II$ (Type of Solution)
$A$. Humidity	$I$. Solid in solid
$B$. Alloys	$II$. Liquid in gas
$C$. Amalgams	$III$. Solid in gas
$D$. Smoke	$IV$. Liquid in solid

The solutions $A$ and $B$ are $0.1 \ M$ and $0.2 \ M$ in a substance. If $100 \ mL$ of $A$ are mixed with $25 \ mL$ of $B$ and there is no change in volume,then the final molarity of the solution is ......... $M$.

Similar Questions

Match List-$I$ with List-$II$.

List-$I$ (Example) List-$II$ (Type of Solution)

$A$. Humidity $I$. Solid in solid

$B$. Alloys $II$. Liquid in gas

$C$. Amalgams $III$. Solid in gas

$D$. Smoke $IV$. Liquid in solid

Choose the correct answer from the options given below:

Dissolving $120 \ g$ of a compound (mol. wt $= 60$) in $1000 \ g$ of water gave a solution of density $1.12 \ g \ mL^{-1}$. The molarity of the solution is: (in $M$)

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$A$ $40 \%$ by mass sucrose solution is heated till it becomes $50 \%$ by mass. Calculate the mass of water lost from $100 \ g$ of the solution. (in $g$)

The normality of $4\% \ (w/V)$ $NaOH$ is

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