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Stuck here, help me understand: - Some basic concepts in chemistry - JEE Main-2

50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hydroxide solution. The amount of NaOH in 50 mL of the given sodium hydroxide solution is :

  • Option 1)

    20 g

  • Option 2)

    4 g

  • Option 3)

    10 g

  • Option 4)

    40 g

 
Answers (1)
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Molar Mass -

The mass of one mole of a substance in grams is called its molar mass.

- wherein

 Molar mass of water = 18 g mol-1

 

 

 

Molarity -

 Molarity (M) = (Number of moles of solute)/(volume of solution in litres)

 

- wherein

It is defined as the number of moles of the solute in 1 litre of the solution.

 

 

 

Stoichiometry -

Stoichiometry deals with measurements of reactants and products in a chemical reaction.

- wherein

aA (g) + bB (g) → cC (g) + dD (g)

Here, ‘a’ moles of A(g) reacts with ‘b’ moles of B(g) to give ‘c’ mole of C(g) and ‘d’ moles of D(g)

As we have learned from mole concept

H_{2}C_{2}O_{4}+2NaOH\rightarrow Na_{2}C_{2}O_{4}+2H_{2}O

meq\, \, \, of H_{2}C_{2}O_{4}=meq\, \, NaOH

50\times 0.5\times 2=25MnaOH\times 1

1000ml\, \, solution= 2\times 240gram NaOH

\therefore 50ml \, \, sol=4g \, \, NaOH

 


Option 1)

20 g

Option 2)

4 g

Option 3)

10 g

Option 4)

40 g

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