Question

Calculate the molarity of 6.3g of oxalic acid $\rm(H_2C_2O_42H_2O)$present in 500 ml of solution.

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Answer 1

Answer:

Molarity of 6.3g of oxalic acid present in 500ml of solution is 0.1 mol/L.

Solution:

Given that,

⇒ Mass of oxalic acid is 6.3g

⇒ It is present in 500ml of solution

We need to find the molarity. Now, let's find the molar mass of oxalic acid.

⇒ C2 = 12×2 = 24

⇒ H2 = 1×2 = 2

⇒ O4 = 16×4 = 64

⇒ 2H2O = 2H2+2O

⇒ 2×1×2+2×16

⇒ 4+32

⇒ 36

Total: 24+2+64+36 = 126g/mol

Hence, molar mass of oxalic acid is 126g/mol. To find the mole, we use the formula:

⇒ Mole = Mass/Molar Mass

⇒ Mole = 6.3/126

⇒ Mole = 0.05mol

Now, to find the molarity, we use the formula:

⇒ Molarity = Number of moles/Volume

⇒ Molarity = 0.05mol/500ml

⇒ Molarity = 0.05mol/0.5l

⇒ Molarity = 0.1 mol/L

Hence, Molarity of 6.3g of oxalic acid present in 500ml of solution is 0.1 mol/L.

Answer 2

hope this equation I did it helps you