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 =