A solution of X2CO3 has a concentration of 15.2 g/dm^3. 25 cm^3 of this solution require 27.5 cm^3 of 0.2 M HCl for complete reaction. Calculate the ralative atomic mass of X
1 dm^3 = 1000 cm^3 = 1 liter
so
X2CO3 is 15.2 g/liter
a liter of this requires 27.5/25 or 1.1 liters of .2 M HCl
1.1 * .2 = .22 mols of HCL
X2CO3 + 2 HCl --> 2 XCl + H2CO3 maybe?
so we only need .11 mols of X2CO3
.11 mols of X2CO3 weigh 15.2 grams
1 mol X2CO3 weighs 138 grams
C is 12
O3 is 48
138 - 12 - 48 = 78
so X2 is 78 so x is 39
(Potassium? K2CO3 )
To calculate the relative atomic mass of X, we need to determine the number of moles of X2CO3 and the number of moles of HCl that react.
First, we can calculate the number of moles of HCl used in the reaction. The concentration of the HCl solution is given as 0.2 M (moles per liter). Since we have 27.5 cm^3 (or 0.0275 dm^3) of the HCl solution, we can calculate the number of moles of HCl as follows:
Moles of HCl = concentration (M) × volume (dm^3)
Moles of HCl = 0.2 M × 0.0275 dm^3
Moles of HCl = 0.0055 mol
According to the balanced chemical equation, 1 mole of X2CO3 reacts with 2 moles of HCl. Therefore, in the given reaction, the number of moles of X2CO3 is twice the number of moles of HCl used:
Moles of X2CO3 = 2 × Moles of HCl
Moles of X2CO3 = 2 × 0.0055 mol
Moles of X2CO3 = 0.011 mol
Now, we need to determine the mass of X2CO3 that corresponds to the calculated number of moles. The concentration of the X2CO3 solution is given as 15.2 g/dm^3. This means that 1 dm^3 of the solution contains 15.2 grams of X2CO3.
Mass of X2CO3 = concentration (g/dm^3) × volume (dm^3)
Mass of X2CO3 = 15.2 g/dm^3 × 0.025 dm^3
Mass of X2CO3 = 0.38 g
Finally, we can calculate the relative atomic mass of X using the equation:
Relative atomic mass of X = Mass of X2CO3 / Moles of X2CO3
Relative atomic mass of X = 0.38 g / 0.011 mol
Relative atomic mass of X ≈ 34.5 g/mol
Therefore, the relative atomic mass of X is approximately 34.5 g/mol.