Answer:
[tex]1.58\ \text{g}[/tex]
Explanation:
10.11 g sample of [tex]NaBr[/tex] contains 22.34% [tex]Na[/tex] by mass
According to the law of constant composition the if one sample of [tex]NaBr[/tex] has 22.34% of [tex]Na[/tex] by mass then any other sample of [tex]NaBr[/tex] will have the same percentage of the amount of [tex]Na[/tex].
For a sample of 7.09 g we have
[tex]7.09\times \dfrac{22.34}{100}=1.583906\approx 1.58\ \text{g}[/tex]
The mass of sodium in the required sample is [tex]1.58\ \text{g}[/tex].
