JEE Exam > JEE Notes > Chapter-wise Tests for JEE Main & Advanced > JEE Advanced (Fill in the Blanks): Some Basic Concepts of Chemistry
JEE Advanced (Fill in the Blanks): Some Basic Concepts of Chemistry | Chapter-wise Tests for JEE Main & Advanced PDF Download
Q.1. The modern atomic mass unit is based on ........................... (1980)
Ans. Carbon-12.
Sol. Carbon-12.
Q.2. The total number of electrons present in 18 ml of water is ....................... (1980)
Ans. 6.02 × 1024
Sol. 6.02 × 1024
18 ml H2O = 18 g H2O (∵ density of water = 1 g/cc)
= 1 mole of H2O.
1 Mole of H2O = 10 × 6.02 × 1023 electrons
(∵ Number of electrons present in one molecule of water
= 2 + 8 = 10)
= 6.02 × 1024 electrons
Q.3. 3 g of a salt of molecular weight 30 is dissolved in 250 g of water. The molality of the solution is ............... . (1983 - 1 Mark)
Ans. 0.4m
Sol. TIPS/Formulae : Molality 
Molality 
= 0.4m
Q.4. The weight of 1 × 1022 molecules of CuSO4.5H2O is ............... (1991 - 1 Mark)
Ans. 4.14 g
Sol. TIPS/Formulae : 1 Mole = 6.023 × 1023 molecules = Molecular weight in gms.
Weight of 6.023 × 1023 (Avogadro’s number) molecules of CuSO4.5H2O = Molecular wt. of CuSO4.5H2O = 249 g.
∴ Weight of 1 × 1022 molecules of CuSO4.5H2O
= 4.14 g
Q.5. The compound YBa 2 Cu3 O7 , which shows super conductivity, has copper in oxidation state..................., assume that the rare earth element yttrium is in its usual + 3 oxidation state. (1994 - 1 Mark)
Ans. 
Sol. NOTE : Sum of oxidation states of all atoms (elements) in a neutral compound is zero.
TIPS/Formulae : As YBa2Cu3O7 is neutral. (+ 3) + 2 (+ 2) + 3 (x) + 7 (–2) = 0 or 3 + 4 + 3x – 14 = 0
⇒ 3x + 7 – 14 = 0 or x =
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