Question

Solution B was formed by dissolving 1.06g of Na2CO3 in 250 cm^3 of solution. Solution A was prepared from a concentrated HCl ( density = 1.18 kg/dm^3, Assay = 36-37% (m/m)) by measuring 0.25 cm^3 of the concentrated HCl in 100 cm^3 of solution. Find the volume of the diluted HCl that is needed to react completely with the Na2CO3

Answer 1

What's the molarity of the HCl solution? Note 1.18 kg/dm^3 = 1.18 g/mL. Also the molarity of the HCl can't be calculated exactly so I'll take the average of 36-37% as the assay. That means the EXACT volume of HCl needed to react with the Na2CO3 isn't known but the value calculated is close.
That's 1000 mL x 1.18 g/mL x 0.365 x (1/36.5) = 11.6 M for the solution A.
Molarity of the diluted HCl solution is 11.6 x (0.25 cc/100 cc) = 0.029 M

Molarity of the Na2CO3 solution.
mols Na2CO3 = grams/molar mass = 1.06/106 = 0.01 mols
M Na2CO3 = mols/L = 0.01/0.250 = 0.04 M

2HCl + Na2CO3 = 2NaCl + H2O + CO2
millimols Na2CO3 = mL x M = 250 mL x 0.04 M = 10
It will take 20 mmols HCl. Look at the coefficients to know that.
M diluted HCl = mmols/mL
0.029M = 20 mmols/mL
Solve for mL = ? but that number looks unrealistic to me. Check out the numbers.