Question

4.
How will you prepare 250 mL of 0.05 M oxalic acid solution ?​

Answer 1

Answer : Molecular formula of solid crystalline oxalic acid: H2C2O4, 2H2O. Its Molecular Wt. 126. Hence, to prepare 250 ml M/20 oxalic acid solution 126*250/(20*1000) = 1.575 gram of oxalic acid is to be dissolved in water in a 250 ml volumetric flask.

Same way prepare for 0.05 M

Answer 2

0.05 M 250 mL of oxalic acid solution will be prepared by adding 1.125 g of oxalic acid to 250 mL of solution.

Given,

Molarity of oxalic acid solution=0.05 M

Volume of oxalic acid solution=250 mL.

To find,

the preparation of 0.05 M oxalic acid solution.

Solution:

• The molecular formula of oxalic acis is (COOH)2.
• The molar mass of (COOH)2 is 90 g.
• Molarity of a substance in a solution is defined as the moles of that substance in 1 L of solution.
• $Molarity=\frac{Moles}{Volume(L)}$
• Molarity is a temperature dependent concentration term.

The molarity of 250 mL oxalic acid solution is 0.05 M.

$Molarity=\frac{Moles}{Volume(L)}\\0.05=\frac{Moles}{\frac{250}{1000} } \\0.05=\frac{Moles}{\frac{1}{4} }\\Moles=\frac{1}{4}X0.05\\Moles=0.0125.$

0.0125 moles of oxalic acid are present in the solution.

$Moles=\frac{Mass}{Molar mass} \\0.0125=\frac{Mass}{90}\\Mass=0.0125X90\\Mass=1.125 g.$

1.125 g of oxalic acid is present in the solution.

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