Suggest a range of sample masses for the indicated primary standard if it is desired to use between 35 and 45mL of titrant:

(a) 0.175 M HClO4 titrated against Na2CO3(CO2 product)
(b) 0.085 M HCl titrated against Na2C2O4
Na2C2O4-->Na2CO3+CO
CO3(2-)+2H(+)-->H2O+CO2
(c) 0.150 M NaOH titrated against benzoic acid
(d) 0.050 M Ba(OH)2 titrated against KH(IO3)2
(e) 0.075 M HClO4 titrated against TRIS
(f) 0.050 M H2SO4 titrated against Na2B4O7 * 10 H20
reaction:
B4O7(2-)+2 H30(+)+3H20-->4H3BO3

a.

2HClO4 + Na2CO3 ==> 2NaCl + 2H2O

mols HClO4 = M x L = 0.035 x 0.175 = approx 0.006125
mols Na2CO3 = 1/2 that from 2 mol HClO4 = 1 mol Na2CO3
mol Na2CO3 = grams/molar mass or grams = mols x molar mass = 0.00306 x 106 = approx 0.320 g.
Now do the same thing with 45 mL instead of 35 mL to get the range. In practice you take approx the average.

To suggest a range of sample masses for a given primary standard, we need to consider the stoichiometry of the reaction and the volume of the titrant to be used.

(a) 0.175 M HClO4 titrated against Na2CO3(CO2 product):
From the reaction given, we can see that 1 mole of HClO4 reacts with 1 mole of Na2CO3. The molarity of HClO4 is 0.175 M. So, for complete reaction, we will need equal moles of HClO4 and Na2CO3.

To determine the mass of Na2CO3 required, we need to use its molar mass. The molar mass of Na2CO3 is approximately 106 g/mol.

Since the volume of the titrant is between 35 and 45 mL, the moles of HClO4 will be between:
0.175 M * 0.035 L = 0.006125 mol
and
0.175 M * 0.045 L = 0.007875 mol

Given that the reaction is stoichiometric, the mass of Na2CO3 required will be between:
0.006125 mol * 106 g/mol = 0.64975 g
and
0.007875 mol * 106 g/mol = 0.83475 g

Therefore, a suggested range of sample masses for Na2CO3 is between 0.65 g and 0.83 g.

You can use the same approach to calculate the sample masses for the remaining cases (b), (c), (d), (e), and (f) by considering the stoichiometry of the reactions and the volumes of titrants.