# A 35 mL sample of concentrated H3PO4 (14.7M) is diluted to a final volume of 250 mL. a) what is the M of the final solution? b) How much solvent is added to the original solution?

Aug 6, 2017

Final concentration.....$\cong 2 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

$a .$ $\text{Molarity"="Moles of solute"/"Volume of solution}$....and thus.....

$\text{Moles of solute"="Molarity"xx"Volume}$.....we gots......

$\frac{35 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 \times 14.7 \cdot m o l \cdot {L}^{-} 1}{250 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1}$

$= 2.058 \cdot m o l \cdot {L}^{-} 1$ $\text{with respect to phosphoric acid.}$

$b .$ We assume, reasonably that the volumes ARE additive; i.e. we started with an aqueous solution of $35 \cdot m L$ volume and we add approx. $215 \cdot m L$ of water to make it up to a $250 \cdot m L$ volume.