`H_2SO_4+2NaOH rarr Na_2SO_4+2H_2O`
Mole ratio;
`NaOH:H_2SO_4 = 2:1`
Amount of `H_2SO_4` consumed `= 0.825/1000xx42.9 = 0.0354mol`
Amount of NaOH reacted `= 2xx0.0354 = 0.0708mol`
If the molarity of NaOH is X;
`X/1000xx75 = 0.0708`
`X = 0.944M`
So the molarity of NaOH is 0.944mol/L
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