For exact neutralization of 200 mL of 0.10 M sodium hydroxide ( $N a O H$ ) solution, what can be the volume of 0.20 molar (M) sulfuric acid (H $_{2}$ SO $_{4}$ ) solution?

400

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50.0

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100

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200

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80

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Solution

The correct option is E $50.0$ mL
$2 N a O H + H_{2} S O_{4} \to N a_{2} S O_{4} + 2 H_{2} O$
Number of moles of NaOH $2 \times$ number of moles of $H_{2} S O_{4}$

$M_{1} V_{1} = 2 \times M_{2} V_{2}$
$M_{1} = 0.10$ M $=$ molarity of $N a O H$
$V_{1} = 200$ mL $=$ volume of $N a O H$
$M_{2} = 0.20$ M $=$ molarity of sulpuric acid.
$V_{2} =$ volume of sulphuric acid

$0.10 \times 200 = 2 \times 0.20 \times V_{2}$
$∴ V_{2} = 50$ mL

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For exact neutralization of 200 mL of 0.10 M sodium hydroxide (NaOH) solution, what can be the volume of 0.20 molar (M) sulfuric acid (H2SO4) solution?

The correct option is E 50.0 mL2NaOH+H2SO4→Na2SO4+2H2ONumber of moles of NaOH 2× number of moles of H2SO4M1V1=2×M2V2M1=0.10 M = molarity of NaOHV1=200 mL = volume of NaOHM2=0.20 M = molarity of sulpuric acid.V2= volume of sulphuric acid0.10×200=2×0.20×V2∴V2=50 mL

For exact neutralization of 200 mL of 0.10 M sodium hydroxide ( $N a O H$ ) solution, what can be the volume of 0.20 molar (M) sulfuric acid (H $_{2}$ SO $_{4}$ ) solution?