Question

СРОЧНО Срочно Срочно Срочно 2 ЗАДАЧА.Визначити молярну концентрацію 6н розчину сульфату залiза Fe2(SO4)3​ 3 ЗАДАЧА. Знайти масову частку (у%) 0,5н розчину H3PO4 (p=1,009 г/моль)​

Answer 1

Hi, I'm more Russian. I need points because I am solving the Olympiad in music, and I can no longer ask the descendant anything here that I have run out of lectures. So sorry

Answer 2

Calculation of Molar Concentration of 6N Iron(III) Sulfate Solution (Fe2(SO4)3)

To determine the molar concentration of a 6N solution of iron(III) sulfate (Fe2(SO4)3), we need to understand the concept of normality (N) and its relationship with molarity (M).

Normality (N) is a measure of the number of equivalents of a solute per liter of solution. It is calculated by multiplying the molarity (M) of the solute by the number of equivalents per mole of the solute.

The molar concentration (M) is the number of moles of solute per liter of solution.

In the case of iron(III) sulfate (Fe2(SO4)3), the compound dissociates into two iron(III) ions (Fe3+) and three sulfate ions (SO4^2-). Therefore, the number of equivalents per mole of Fe2(SO4)3 is 5 (2 equivalents of Fe3+ and 3 equivalents of SO4^2-).

To calculate the molar concentration (M) of the 6N solution of Fe2(SO4)3, we can use the following formula:

Molar Concentration (M) = Normality (N) / Number of Equivalents per Mole

For a 6N solution, the normality (N) is 6. Therefore, we can substitute the values into the formula:

M = 6 / 5

Now, let's calculate the molar concentration:

M = 1.2 M

So, the molar concentration of the 6N solution of iron(III) sulfate (Fe2(SO4)3) is 1.2 M.

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