Question

what is the molarity of the resulting solution when 300. ml of a 0.600 m aqueous solution is diluted by the addition of 400. ml of water? assume volumes are additive.
0.257 m
0.450 m
0.800 m
1.40 m

Explanation:

Step1: Calculate the moles of solute

Use the formula \(n = M_1V_1\). Given \(M_1=0.600\space M\) and \(V_1 = 300\space mL=0.300\space L\).
\(n=(0.600\space M)\times(0.300\space L) = 0.180\space mol\)

Step2: Calculate the final volume

\(V_2=(300 + 400)\space mL=700\space mL = 0.700\space L\)

Step3: Calculate the final molarity

Use the formula \(M_2=\frac{n}{V_2}\). Substitute \(n = 0.180\space mol\) and \(V_2=0.700\space L\).
\(M_2=\frac{0.180\space mol}{0.700\space L}\approx0.257\space M\)

Answer:

0.257 M