Question

How many molecules of water of hydration are present in 630 mg of oxalic acid. Molecular formula of oxalic acid is H2 C2 04. 2H20

Answer 1

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Answer 2

Answer: The number of molecules of water in given amount of oxalic acid is $6.022\times 10^{21}$

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of oxalic acid = 630 mg = $630\times 10^{-3}g$   (Conversion factor: 1 g = 1000 mg)

Molar mass of oxalic acid = 126 g/mol

Putting values in above equation, we get:

$\text{Moles of oxalic acid}=\frac{630\times 10^{-3}g}{126g/mol}=0.005mol$

Chemical formula fr oxalic acid is $C_2H_2O_4.2H_2O$

By Stoichiometry:

1 mole of oxalic acid contains $2\times N_A=2\times 6.022\times 10^{23}$ number of water molecules.

Thus, 0.005 moles of oxalic acid will contain $0.005\times 2\times 6.022\times 10^{23}=6.022\times 10^{21}$ number of water molecules.

Hence, the number of molecules of water in given amount of oxalic acid is $6.022\times 10^{21}$