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a 0.3503 g sample of an unknown diprotic acid (h2x) was dissolved in 50.0ml of water and titrated with the same lithium hydroxide solution as in question 15. if it took 26.60 ml of the lithium hydroxide to reach neutralize the acid, what was the molar mas of the unknown acid?

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In this problem, we can use the titration data to determine the amount of lithium hydroxide (LiOH) that reacted with the unknown diprotic acid (H2X), and then use this information to calculate the molar mass of the acid.

First, we need to calculate the number of moles of LiOH that were added to the solution:

moles of LiOH = Molarity of LiOH * volume of LiOH used in liters

From question 15, we know that the molarity of the LiOH solution is 0.1000 M. Converting the volume of LiOH used from milliliters to liters:

volume of LiOH used = 26.60 mL = 0.02660 L

So, the moles of LiOH used in the titration is:

moles of LiOH = 0.1000 M * 0.02660 L = 0.002660 moles

Since the diprotic acid is being titrated with a strong base, each mole of LiOH reacts with two moles of H2X. Therefore, the number of moles of H2X can be calculated as:

moles of H2X = 0.002660 moles LiOH / 2 = 0.001330 moles H2X

Next, we can calculate the molar mass of the unknown diprotic acid using its mass and the number of moles:

molarmass of H2X = mass of H2X / moles of H2X

The mass of the unknown diprotic acid is given as 0.3503 g, so:

molar mass of H2X = 0.3503 g / 0.001330 moles = 262.8 g/mol

Therefore, the molar mass of the unknown diprotic acid is 262.8 g/mol.

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