32.In a titration experiment, a student was provided with the following:-
− Solution R which was 0.408M HCl
− Solution S containing 6.9g of a metal carbonate M2CO3 in 250cm3
solution
Upon titration, 24.5cm3 of solution R was required to completely neutralize 25cm3 of solution S.
Determine the following.
a. Molarity of solution S
b. Formula mass of M2CO3
c. Relative atomic mass of M
I m interested in part b and C i want to get clear workings
To determine the formula mass of M2CO3 and the relative atomic mass of M, we can follow these steps:
a. Molarity of solution S:
First, we need to determine the number of moles of HCl in solution R that reacted with solution S.
Moles of HCl = concentration x volume in liters
Moles of HCl = 0.408 mol/L x 24.5 cm3 / 1000 cm3/L = 0.00999 moles
Since the reaction is a 1:1 ratio between HCl and the metal carbonate M2CO3, the number of moles of M2CO3 in solution S is also 0.00999 moles.
Molarity of solution S = moles / volume in liters
Molarity of solution S = 0.00999 moles / 0.025 L = 0.3996 M
b. Formula mass of M2CO3:
To find the formula mass of M2CO3, we need to find the molar mass of M2CO3 by adding up the molar masses of the elements in the compound.
From the reaction, we know that 0.00999 moles of M2CO3 reacted with HCl.
Molar mass of M2CO3 = mass / moles
Mass of M2CO3 = 6.9g
Molar mass of M2CO3 = 6.9g / 0.00999 mol = 690 g/mol
c. Relative atomic mass of M:
The formula mass of M2CO3 is the sum of the atomic masses of M, C, and O in the compound.
Formula mass of M2CO3 = 690g/mol
Molar mass of C (carbon) = 12.01 g/mol
Molar mass of O (oxygen) = 16.00 g/mol
To find the relative atomic mass of M, we can rearrange the formula mass equation:
690 g/mol = (molar mass of M) + 2(12.01 g/mol) + 3(16.00 g/mol)
690 g/mol = molar mass of M + 24.02 g/mol + 48.00 g/mol
Molar mass of M = 690 g/mol - 72.02 g/mol
Molar mass of M = 617.98 g/mol
Therefore, the relative atomic mass of M is 617.98 g/mol.