1 gram of sodium hydroxide was treated with 25 mL of 0.75 M HCl solution, the mass of sodium hydroxide left unreacted is equal to

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Correct answer: B. (b) NaOH+ HCl ⟶ NaCl+H_2O Given ⇒1 g (0.75M ,25 mL ) Mole of NaOH= 140 mole =0.025 mol Mole of HCl= 0.75 251000=0.01875 mol Now finding the L.R Value = Given AmountS .C NaOH : HCl 0.0251> 0.018751 So, L.R will be HCl Unreacted moles of NaOH =0.025-0.01875 =0.00625 mol Unreacted mass =0.00628 40 g =0.25 g=250 mg

$1 gram of sodium hydroxide was treated with 25 mL of 0.75 M HCl solution, the mass of sodium hydroxide left unreacted is equal to$

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