dissolving 17.16 g hydrated sodium carbonate [ Na2CO3.XH2O] in water and complete the volume to 500 mL , if 25 mL of this solution neutralize with 30 mL of 0.2 M hydrochloric acid, Calculate the percentage of water in hydrated salt. [ Na= 23 , C = 12 , O = 16 H =1 ]a) 33.75%b) 35.41%c) 45.67%d) 62.94%

To calculate the percentage of water in the hydrated sodium carbonate, we need to determine the number of moles of hydrochloric acid that reacted with the sodium carbonate.

First, let's calculate the number of moles of hydrochloric acid used. We know that the volume of the hydrochloric acid is 30 mL and the concentration is 0.2 M.

Number of moles of hydrochloric acid = volume (in L) x concentration = 0.03 L x 0.2 M = 0.006 moles

According to the balanced chemical equation, 1 mole of hydrochloric acid reacts with 1 mole of sodium carbonate. Therefore, the number of moles of sodium carbonate used is also 0.006 moles.

Next, we need to calculate the molar mass of hydrated sodium carbonate. The molar mass of Na2CO3 is (2 x 23) + 12 + (3 x 16) = 106 g/mol.

Now, let's calculate the mass of hydrated sodium carbonate used. We know that the mass of the hydrated sodium carbonate is 17.16 g.

Number of moles of hydrated sodium carbonate = mass / molar mass = 17.16 g / 106 g/mol = 0.162 moles

Since the number of moles of sodium carbonate used is the same as the number of moles of hydrated sodium carbonate used, we can conclude that the molar ratio of water to sodium carbonate is 1:1.

Therefore, the number of moles of water in the hydrated sodium carbonate is also 0.162 moles.

Now, let's calculate the mass of water in the hydrated sodium carbonate. The molar mass of water is 18 g/mol.

Mass of water = number of moles x molar mass = 0.162 moles x 18 g/mol = 2.916 g

Finally, let's calculate the percentage of water in the hydrated salt.

Percentage of water = (mass of water / mass of hydrated salt) x 100 = (2.916 g / 17.16 g) x 100 = 16.98%

Therefore, the percentage of water in the hydrated salt is approximately 16.98%.