# Solved Examples on Some Basic Concepts of Chemistry

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)

1) C3H4

2) C6H5

3) C7H8

4) C2H4

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,

M1V1 = 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)

1) 1.78 M
2) 2.00 M
3) 2.05 M
4) 2.22 M

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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