# Question 55b54

Nov 6, 2017

$2.1 {\mathrm{dm}}^{3}$

#### Explanation:

$12 c {m}^{3} = 0.012 {\mathrm{dm}}^{3}$ of "0.98 mol dm^-3 H_2 SO _4

## 1

Will use the useful formula:

$C = \frac{n}{V} \to 0.98 = \frac{n}{0.012} \to n = 0.1176 m o l$

Then we will use formula again to know how much volume of the first solution will need to prepare the second:

$C = \frac{n}{V} \to 0.056 = \frac{0.1176}{V} \to V = 2.1 {\mathrm{dm}}^{3}$

$2$

Ww will use the other expression:

$C 1 \cdot V 1 = C 2 \cdot V 2$

$0.98 \cdot 0.012 = 0.058 \cdot V 2 \to V 2 = 2.1 {\mathrm{dm}}^{3}$