# What is the new concentration of a solute when a 20*mL volume of a mother liquor at 2.0*mol*L^-1 concentration is diluted to a volume of 100*mL?

Aug 20, 2017

The new concentration is $0.40 \cdot m o l \cdot {L}^{-} 1$.
$\text{Molarity"="Moles of solute"/"Volume of solution}$
Initially, we have a molar quantity of $20 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 \times 2 \cdot m o l \cdot {L}^{-} 1 = 0.040 \cdot m o l$, and this molar quantity is further diluted to $100 \cdot m L$....
$\text{Concentration} = \frac{0.040 \cdot m o l}{100 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1} = 0.40 \cdot m o l \cdot {L}^{-} 1$.