Figure shows a manometer, one arm is connected with a bulb containing ${NH}_{3}$ and other arm containing $N_{2} H_{4}$ both at initial pressure of 1 atm. Initially both arms of manometer contains liquid at same level, What will be difference in 'level of arms, when 50% ${NH}_{3}$ (g) and 75% of $N_{2} H_{4}$ (g) dissociates according to the given reaction. Assume temperature remains constant?
$\begin{aligned} 2 {NH}_{3} (g) ⟶ N_{2} (g) + 3 H_{2} (g) \\ N_{2} H_{4} (g) ⟶ N_{2} (g) + 2 H_{2} (g) \end{aligned}$

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38 cm if liquid having density 27.2 g/mL is used

152 cm if liquid having density 6.8 g/mL is used

304 cm if liquid having density 3.4 g/mL is used

76 cm if liquid having density 13.6 g/mL is used

answer is A, B, C, D.

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Detailed Solution

$\begin{array}{cccccc} 2 {NH}_{3} & ⟶ N_{2} + 3 H_{2}; & N_{2} H_{4} & ⟶ & N_{2} + 2 H_{2} \\ 1 - P & \frac{P}{2} & \frac{3 P}{2} & 1 - x & x & 2 x \\ 0.5 & 0.25 & 0.75 & 0.25 & 0.75 & 1.5 \\ Total = 1.5 & Total = 2.5 \end{array}$

Difference in pressure = 1 atm = 76 cm of Hg

$\begin{aligned} 76 \times 13.6 = h_{1} \times 27.2 = h_{2} \times 6.8 = h_{3} \times 3.4 \\ \Rightarrow h_{1} = 38 cm; h_{2} = 152 cm; h_{3} = 304 cm \end{aligned}$