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 :
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
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