The amount of Zn (atomic weight = 65) required to form 224 c.c. of H₂ at N.T.P. on treatment with dilute H₂SO₄ will be __________ gm.

ANS:B - 0.65

To find the amount of zinc (Zn) required to produce hydrogen gas (H2) at NTP (Normal Temperature and Pressure), we need to consider the stoichiometry of the reaction between zinc and dilute sulfuric acid (𝐻2𝑆𝑂4H2​SO4​). The reaction is: Zn+H2SO4→ZnSO4+H2Zn+H2​SO4​→ZnSO4​+H2​ From the balanced equation, we can see that 1 mole of zinc (ZnZn) produces 1 mole of hydrogen gas (H2H2​). At NTP, 1 mole of any gas occupies 22.4 liters or 22.4×10322.4×103 cc. Given that 224 cc of hydrogen gas is produced, it means 22422.4×10322.4×103224​ moles of zinc is required. The atomic weight of zinc is 65 g/mol. Therefore, the amount of zinc required in grams is: Amount (g)=Moles×Atomic weightAmount (g)=Moles×Atomic weight Amount (g)=22422.4×103×65Amount (g)=22.4×103224​×65 Amount (g)=0.65Amount (g)=0.65 So, the correct answer is 0.650.65 gm. Therefore, the answer is Option 2: 0.650.65