8.58g of coastic soada was dissolved to produce a solution of 500cm3 .to neutralize 50cm3 of this solution 15cm3 of 0.4M HCl was consumed .(coastic soada = Na2CO3.XH2O ) find the value of X?
Note the correct spelling of caustic soda. Also, correct sentence structure helps us read these problems better.
How many mols HCl did you use? That's M x L = 0.4M x 0.015L = 0.006.
How many grams were in the solution you titrated. That's 8.58 x 50/500 = 0.858 g.
Then mols = grams/molar mass and rearrange to molar mass = grams/mols.
0.858/0.006 = about 143.
Subtract Na2CO3 = -106
Remainder is 37 and 37/18 = 2.05 mols and I would round that to 2.0 for X.
To find the value of X in Na2CO3·XH2O, we first need to know the molar ratio between Na2CO3 and HCl.
From the given information:
• Volume of the Na2CO3 solution = 500 cm3
• Volume of HCl consumed = 15 cm3
• Concentration of HCl = 0.4 M
We can calculate the moles of HCl consumed using the formula:
moles of HCl = concentration (M) × volume (L)
First, convert the volume of HCl consumed to liters:
15 cm3 = 15 cm3 ÷ 1000 cm3/L = 0.015 L
Now, substitute the values into the formula:
moles of HCl = 0.4 M × 0.015 L = 0.006 moles
Since HCl reacts with Na2CO3 in a 1:1 molar ratio, we can conclude that 0.006 moles of Na2CO3 also reacted.
Next, we need to calculate the number of moles present in the 500 cm3 of the Na2CO3 solution.
moles of Na2CO3 in 500 cm3 = concentration (M) × volume (L)
Here, the concentration of Na2CO3 is not given directly, but we can calculate it by:
concentration (M) = mass (g) ÷ molar mass (g/mol)
Given that 8.58 g of Na2CO3 was dissolved to produce the 500 cm3 solution:
molar mass of Na2CO3 = 2 × atomic mass of Na + atomic mass of C + 3 × atomic mass of O
molar mass of Na2CO3 = 2 × 22.99 g/mol + 12.01 g/mol + 3 × 16.00 g/mol
= 105.99 g/mol
Substitute the values into the formula:
concentration of Na2CO3 = 8.58 g ÷ 105.99 g/mol
= 0.081 M
Now, calculate the moles of Na2CO3 in the 500 cm3 of the solution:
moles of Na2CO3 in 500 cm3 = 0.081 M × 0.5 L
= 0.0405 moles
Since we know that 0.006 moles of Na2CO3 reacted, we can subtract this amount to find the moles of water (XH2O) present in Na2CO3·XH2O:
moles of H2O = moles of Na2CO3 in the solution - moles of Na2CO3 reacted
= 0.0405 moles - 0.006 moles
= 0.0345 moles
Finally, to find the value of X, we need to calculate the ratio between the moles of water and Na2CO3:
X = moles of H2O ÷ moles of Na2CO3
= 0.0345 moles ÷ 0.0405 moles
≈ 0.852
Therefore, the value of X in Na2CO3·XH2O is approximately 0.852.