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

Vedclass · Accepted Answer

$0.12$

Vedclass · Answer

(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 $M_{final} = \frac{n_{total}}{V_{total}} = \frac{15 \ mmol}{125 \ mL} = 0.12 \ M$.

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 ......... $M$.

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