Certainly! Let's break down the question and solve it step-by-step.
We are given:
1. The volume of the [tex]\( \text{H}_2\text{SO}_4 \)[/tex] solution is [tex]\( 300 \, \text{cm}^3 \)[/tex].
2. The mass concentration of the [tex]\( \text{H}_2\text{SO}_4 \)[/tex] solution, which is [tex]\( 10^{-4} \, \text{g/dm}^3 \)[/tex].
The goal is to find the amount of [tex]\( \text{H}_2\text{SO}_4 \)[/tex] in grams contained in [tex]\( 300 \, \text{cm}^3 \)[/tex] of the solution.
### Step 1: Convert the volume from [tex]\( \text{cm}^3 \)[/tex] to [tex]\( \text{dm}^3 \)[/tex]
The relationship between [tex]\( \text{cm}^3 \)[/tex] and [tex]\( \text{dm}^3 \)[/tex] is:
[tex]\[ 1 \, \text{dm}^3 = 1000 \, \text{cm}^3 \][/tex]
So, we convert the given volume:
[tex]\[ 300 \, \text{cm}^3 = \frac{300}{1000} \text{dm}^3 \][/tex]
[tex]\[ = 0.3 \, \text{dm}^3 \][/tex]
### Step 2: Calculate the amount of [tex]\( \text{H}_2\text{SO}_4 \)[/tex] in grams
To find the mass of [tex]\( \text{H}_2\text{SO}_4 \)[/tex] in the solution, we use the mass concentration provided:
[tex]\[ \text{Mass concentration} = 10^{-4} \, \text{g/dm}^3 \][/tex]
Since we have [tex]\( 0.3 \, \text{dm}^3 \)[/tex] of the solution, we can multiply the volume by the mass concentration to find the mass of [tex]\( \text{H}_2\text{SO}_4 \)[/tex]:
[tex]\[ \text{Mass of } \text{H}_2\text{SO}_4 = \text{Volume} \times \text{Mass concentration} \][/tex]
Substituting the values:
[tex]\[ \text{Mass of } \text{H}_2\text{SO}_4 = 0.3 \, \text{dm}^3 \times 10^{-4} \, \text{g/dm}^3 \][/tex]
[tex]\[ = 0.3 \times 10^{-4} \, \text{g} \][/tex]
[tex]\[ = 3 \times 10^{-5} \, \text{g} \][/tex]
Therefore, the amount of [tex]\( \text{H}_2\text{SO}_4 \)[/tex] in [tex]\( 300 \, \text{cm}^3 \)[/tex] of the solution is [tex]\( 3 \times 10^{-5} \, \text{g} \)[/tex].
To summarize:
- The volume of the solution in [tex]\( \text{dm}^3 \)[/tex] is [tex]\( 0.3 \, \text{dm}^3 \)[/tex].
- The amount of [tex]\( \text{H}_2\text{SO}_4 \)[/tex] in grams is [tex]\( 3 \times 10^{-5} \, \text{g} \)[/tex].
