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JEE Main Mock Test Series 2020 & Previous Year Papers

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Question 1: A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g of CO2. The empirical formula of the hydrocarbon is:    (IIT JEE -2013)
(a) C3H4
(b) C6H5
(c) C7H8
(d) C2H4
Answer: c
Solution:
General equation for combustion of hydrocarbon:
CxHy + (x+ y/4)O2 → xCO2 + (y/2)H2O
Number of moles of CO2 produced = 3.08/44 = 0.07
Number of moles of H2O produced = 0.72/18 = 0.04
SO, x / (y/2) = 0.07/0.04 =  7/4
The formula of hydrocarbon is C7H8
Hence, the correct option is c.

Question 2: 29.2 % (w/w) HCl stock solution has density of 1.25 g mL1. The molecular weight of HCl is 36.5 g mol-1. The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is    (IIT JEE 2012)
Solution:
Let mass of the stock solution = 100g
Mass of HCl in 100 g of 29.2 % (w/w) HCl stock solution = 29.2 g
Volume of the stock solution = 100g/1.25 g mL1 = 80 g
Number of moles of HCl in stock solution = 29.2/36.5 = 0.8
Molarity of the stock solution = (0.82/80)×1000 = 10 M
Using,
M1V= M2V2
10× V1 =0.4×200
or
V1 = 8mL
Hence, the volume required is 8 mL.

Question 3: Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density 1.15 g/mL. The molarity of the solution is    (IIT JEE-2011)
(a) 1.78 M
(b) 2.00 M
(c) 2.05 M
(d) 2.22 M
Answer: c
Solution:
For calculating molarity of the solution we require to know two things, (i) number of moles of urea & (ii) total volume of the solution.
Number of moles of urea dissolved in the solution = 120 g/60g = 2.
Total mass of the solution = 1000g + 120 g = 1120 g
Volume of the total solution = 1120g/1.15 gmL-1 = 974 mL = 0.974 L
Molarity = 2/0.974 = 2.05 M
Hence, the correct option is c.

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