# If 10 grams of AgNO3 is available, what volume of 0.25 M AgNO3 solution can be prepared?

Sep 20, 2016

Approx. $0.25 \cdot L$, but you should make a better approximation

#### Explanation:

$\text{Moles of silver nitrate}$ $=$ $\frac{10 \cdot g}{169.87 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.0589 \cdot m o l$.

We require $0.25 \cdot m o l \cdot {L}^{-} 1$ $A g N {O}_{3} \left(a q\right)$, i.e.

$\frac{0.0589 \cdot m o l}{\text{Volume of solution}}$ $=$ $0.25 \cdot m o l \cdot {L}^{-} 1$

Or $\text{Volume of solution}$ $=$ $\frac{0.0589 \cdot \cancel{m o l}}{0.25 \cdot \cancel{m o l} \cdot {L}^{-} 1}$ $\cong$ $0.25 \cdot L$