Question

X is a solution,of dibasic of H2X. G is a solution containing 1.00g of NaOH in 250cm3 of solution . 24.60cm3 of solution F required 25.00cm3 of solution G . From your result and information provided above. (I)concentration of G in mol/dm3 (ii) concentration of F in mol/dm3 (iii) molar mass. Of acid H2X given that 100cm3 of solution F contained 0.495g of acid (H-1.00,O-6,Na-23)

Answer 1

You made a typo. O is 16 not 6.
i) concn G is 0.1 mol/dm3. Calculated as follows:
mols NaOH = g/molar mass = 0.1/40 = 0.025
M NaOH = 0.025/0.25 = 0.1 mol/dm3

ii)Note: 1 L = 1 dm3 and M = mols/dm3
2NaOH + H2X ==> Na2X + 2H2O
mols NaOH used = M x L = 0.1M x 0.025 = 0.0025
mols H2X (F) = 1/2 x 0.0025 = 0.00125
M F = mols/L = 0.00125/0.0246 = 0.051M = 0.051 mol/dm3 but you may want the answer more accurate than that.

iii)
mols H2X in 100 cc = M x L = 0.051 x 0.1 = 0.0051
mols = grams/molar mass or
molar mass = grams/mols = 0.495/0.00508 = ?

Answer 2

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