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### JEE Main 2016 (Online) 9th April Morning Slot

The amount of arsenic pentasulphide that can be obtained when 35.5 g arsenic acid istreated with excess H2S in the presence of conc. HCl ( assuming 100% conversion) is :
A
0.50 mol
B
0.25 mol
C
0.125 mol
D
0.333 mol

## Explanation

2H3 As O4 + 5H2S  $\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{HCl}^{conc}}$   As2S5 + 8H2O

35.5 g of H3AsO4 = ${{35.5} \over {142}}$ = 0.25 moles

Let, As2S5 produced = n moles.

$\therefore\,\,\,$ ${{0.25} \over 2} = {n \over 1}$

$\Rightarrow $$\,\,\, n = 0.125 mol. 2 MCQ (Single Correct Answer) ### JEE Main 2016 (Online) 9th April Morning Slot An organic compound contains C, H and S. The minimum molecular weight of the compound containing 8% sulphur is : (atomic weight of S = 32 amu) A 200 g mol-1 B 400 g mol-1 C 600 g mol−1 D 300 g mol−1 3 MCQ (Single Correct Answer) ### JEE Main 2016 (Online) 9th April Morning Slot 5 L of an alkane requires 25 L of oxygen for its complete combustion. If all volumes are measured at constant temperature and pressure, the alkane is : A Ethane B Propane C Butane D Isobutane ## Explanation We know, conbustion of Hydrocarbon Cx Hy + (x + {y \over 4})O2 \to x CO2 + {y \over 2} H2O So, to consume 1 mole of Cx Hy we need (x + {y \over 4}) mole of O2 gas. As both Cx Hy and O2 are gas, So avogadro's law is applicable on them. At constant temperature and pressure according to avogadro's law, volume \propto mole So, for 5L of Cx Hy the volume of O2 = 5(x + {y \over 4}) According to the question, 5(x + {y \over 4}) = 25 \therefore\,\,\, (x + {y \over 4}) = 5 . . . . . (1) For Ethane (C2 H6), x = 2 and y = 6 \therefore\,\,\, x + {y \over 4} = 2 + {6 \over 4} \ne 5 \therefore\,\,\, C2 H6 can't be the required alkane. For Propane (C3 H8), x = 3. and y = 8 \therefore\,\,\, x + {y \over 4} = 3 + {8 \over 4} = 5 \therefore\,\,\, Propane (C3 H8) is the right alkane. Similarly for Butane (C4 H10) and Isobutane (C4 H10) you can check x + {y \over 4} \ne 5. 4 MCQ (Single Correct Answer) ### JEE Main 2016 (Online) 10th April Morning Slot The volume of 0.1N dibasic acid sufficient to neutralize 1 g of a base that furnishes 0.04 mole of OH in aqueous solution is : A 200 mL B 400 mL C 600 mL D 800 mL ## Explanation According to law of equivalence, Equivalence of acid = Equivalence of base. Equivalence of acid = Normality x volume = 0.1 \times v As we know base produce OH- ion, so moles of base is same as moles of OH- ion = 0.04 Another formula of equivalence = n factor \times number of moles \therefore\,\,\, Equivalance of base = n factor of OH- \times moles of OH- = 1 \times 0.04 As for any ion, the charge of that ion is the n factor of that ion. Here OH- has 1 negative charge so it's n factor = 1 \therefore\,\,\, 0.1 \times v = 1 \times 0.04 \Rightarrow$$\,\,\,$ v = 0.4 L

=   0.4 $\times$ 1000

=    400 ml.