Respuesta :
Answer: The molarity of [tex]Cr^{3+}[/tex] ions in the solution is 0.299 M
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex] .....(1)
- For chromium (III) acetate:
Molarity of chromium (III) acetate solution = 0.234 M
Volume of solution = 26.2 mL
Putting values in equation 1, we get:
[tex]0.234=\frac{\text{Moles of chromium (III) acetate}\times 1000}{26.2}\\\\\text{Moles of chromium (III) acetate}=\frac{0.234\times 26.2}{1000}=0.00613mol[/tex]
1 mole of chromium (III) acetate [tex](Cr(CH_3COO)_3)[/tex] produces 1 mole of chromium [tex](Cr^{3+})[/tex] ions and 3 moles of acetate [tex](CH_3COO^-)[/tex] ions
Moles of [tex]Cr^{3+}\text{ ions}=(1\times 0.00613)=0.00613moles[/tex]
- For chromium (III) nitrate:
Molarity of chromium (III) nitrate solution = 0.461 M
Volume of solution = 10.7 mL
Putting values in equation 1, we get:
[tex]0.461=\frac{\text{Moles of chromium (III) nitrate}\times 1000}{10.7}\\\\\text{Moles of chromium (III) nitrate}=\frac{0.461\times 10.7}{1000}=0.00493mol[/tex]
1 mole of chromium (III) nitrate [tex](Cr(NO_3)_3)[/tex] produces 1 mole of chromium [tex](Cr^{3+})[/tex] ions and 3 moles of nitrate [tex](NO_3^-)[/tex] ions
Moles of [tex]Cr^{3+}\text{ ions}=(1\times 0.00493)=0.00493moles[/tex]
- For chromium cation:
Total moles of chromium cations = [0.00613 + 0.00493] = 0.01106 moles
Total volume of solution = [26.2 + 10.7] = 36.9 mL
Putting values in equation 1, we get:
[tex]\text{Molarity of }Cr^{3+}\text{ cations}=\frac{0.01106\times 1000}{36.9}\\\\\text{Molarity of }Cr^{3+}\text{ cations}=0.299M/tex]
Hence, the molarity of [tex]Cr^{3+}[/tex] ions in the solution is 0.299 M
