50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hydroxide solution. The amount of in 50 mL of the given sodium hydroxide solution is :
20 g
4 g
10 g
40 g
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
Option 1)
20 g
Option 2)
4 g
Option 3)
10 g
Option 4)
40 g
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