Q.51. What volume of hydrogen gas at STP would be liberated by action of 50 mL of of 50% purity (de

Q: What is the correct answer for JEE Main Chemistry Q51?

The correct answer is C. First, calculate the mass of pure $H_2SO_4$: Volume of $H_2SO_4$ solution = 50 mL Density of solution = 1.3 g/mL Mass of $H_2SO_4$ solution = Volume $\times$ Density = 50 mL $\times$ 1.3 g/mL = 65 g. Purity = 50%, so mass of pure $H_2SO_4 = 65 \text{ g} \times 0.50 = 32.5 \text{ g}$. Now, calculate the moles of $H_2SO_4$: Molar mass of $H_2SO_4 = 2(1) + 32 + 4(16) = 2 + 32 + 64 = 98$ g/mol. Moles

Q: Q.51. What volume of hydrogen gas at STP would be liberated by action of 50 mL of of 50% purity (density = 1.3 g mL⁻¹) on 20 g of zinc ? Given : Molar mass of H, O, S, Zn are 1, 16, 32, 65 g mol⁻¹…

The correct answer is C. First, calculate the mass of pure $H_2SO_4$: Volume of $H_2SO_4$ solution = 50 mL Density of solution = 1.3 g/mL Mass of $H_2SO_4$ solution = Volume $\times$ Density = 50 mL $\times$ 1.3 g/mL = 65 g. Purity = 50%, so mass of pure $H_2SO_4 = 65 \text{ g} \times 0.50 = 32.5 \text{ g}$. Now, calculate the moles of $H_2SO_4$: Molar mass of $H_2SO_4 = 2(1) + 32 + 4(16) = 2 + 32 + 64 = 98$ g/mol. Moles

Q: What are the options for this JEE Main Chemistry question?

A. 824 L · B. 428 L · C. 892 L · D. 375 L

Question

Q.51. What volume of hydrogen gas at STP would be liberated by action of 50 mL of  of 50% purity (density = 1.3 g mL⁻¹) on 20 g of zinc ? Given : Molar mass of H, O, S, Zn are 1, 16, 32, 65 g mol⁻¹ respectively.

FirstInTest · Accepted Answer

First, calculate the mass of pure $H_2SO_4$:
Volume of $H_2SO_4$ solution = 50 mL
Density of solution = 1.3 g/mL
Mass of $H_2SO_4$ solution = Volume $	imes$ Density = 50 mL $	imes$ 1.3 g/mL = 65 g.
Purity = 50%, so mass of pure $H_2SO_4 = 65 	ext{ g} 	imes 0.50 = 32.5 	ext{ g}$.

Now, calculate the moles of $H_2SO_4$:
Molar mass of $H_2SO_4 = 2(1) + 32 + 4(16) = 2 + 32 + 64 = 98$ g/mol.
Moles of $H_2SO_4 = \frac{	ext{mass}}{	ext{molar mass}} = \frac{32.5 	ext{ g}}{98 	ext{ g/mol}} \approx 0.3316 	ext{ mol}$.

Mass of zinc = 20 g.
Molar mass of zinc (Zn) = 65 g/mol.
Moles of zinc = $\frac{20 	ext{ g}}{65 	ext{ g/mol}} \approx 0.3077 	ext{ mol}$.

The balanced chemical reaction is:
$Zn(s) + H_2SO_4(aq) ightarrow ZnSO_4(aq) + H_2(g)$
From the stoichiometry, 1 mole of Zn reacts with 1 mole of $H_2SO_4$ to produce 1 mole of $H_2$.

Compare moles of reactants:
Moles of Zn = 0.3077 mol
Moles of $H_2SO_4$ = 0.3316 mol
Since moles of Zn (0.3077 mol) are less than moles of $H_2SO_4$ (0.3316 mol), zinc is the limiting reactant.

Therefore, the moles of $H_2$ produced will be equal to the moles of Zn reacted, which is 0.3077 mol.

At STP (Standard Temperature and Pressure), 1 mole of any ideal gas occupies 22.7 L (according to IUPAC definition, $0^\circ C$ and 1 bar). If using old definition ($0^\circ C$ and 1 atm), it's 22.4 L. Given the options, 22.7 L is more likely.
Volume of $H_2$ at STP = Moles of $H_2 	imes$ Molar volume at STP
Volume of $H_2 = 0.3077 	ext{ mol} 	imes 22.7 	ext{ L/mol} \approx 6.99 	ext{ L}$.

This is closest to 6.892 L (Option 3). The solution uses $20/65 	imes 22.7 = 6.9 L$. This confirms the use of 22.7 L/mol for STP.

Answer

824 L

Answer

428 L

Answer

892 L

Answer

375 L

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