Ans: One mole of a substance is the amount of substance that has the same number of particles as the number of carbon atoms in 12g of C-12 carbon. In 1 mole of a substance there are 6.023 x 10²³ number of atoms.
To calculate the number of molecules in a substance, find the molecular weight of the substance. This molecular weight will have the avogadro number of molecules. Now calculate the molecules for the given mass of the substance.
In the question:
Molecular weight of H₂O is 18g
18g of H₂O = 6.023 x 10²³ molecules
Now, 34g of H₂O = 6.023 x 10²³ x 34/18
= 11.33 x 10²³ molecules
Molecular weight of CO₂ is 44 g
44 g of CO₂ = 6.023 x 10²³ molecules
Now, 28g of CO₂ = 6.023 x 10²³ x 28/44
=3.8 x 10²³ molecules
Molecular weight of ethyl alcohol = 32g
32 g of ethyl alcohol = 6.023 x 10²³ molecules
Now, 46 g of ethyl alcohol = 6.023 x 10²³ x 46/32
= 8.625 x10²³ molecules
Molecular weight of N₂O₅ = 108g
108 g of N₂O₅ = 6.023 x 10²³ molecules
Now, 56 g of N₂O₅ = 6.023 x 10²³ x 56/108
= 3 x 10²³ molecules
Therefore, 34g of water has the highest number of molecules.
Brass alloy is made of which metals?
The reaction sequence given below is carried out with 16 moles of X. The yield of the major product in each step is given below the product in parentheses. The amount (in grams) of S produced is ____. 
Use: Atomic mass (in amu): H = 1, C = 12, O = 16, Br = 80
Let $ a_0, a_1, ..., a_{23} $ be real numbers such that $$ \left(1 + \frac{2}{5}x \right)^{23} = \sum_{i=0}^{23} a_i x^i $$ for every real number $ x $. Let $ a_r $ be the largest among the numbers $ a_j $ for $ 0 \leq j \leq 23 $. Then the value of $ r $ is ________.
Let $ \mathbb{R} $ denote the set of all real numbers. Then the area of the region $$ \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x},\ 5x - 4y - 1 > 0,\ 4x + 4y - 17 < 0 \right\} $$ is
The center of a disk of radius $ r $ and mass $ m $ is attached to a spring of spring constant $ k $, inside a ring of radius $ R>r $ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following Hooke’s law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $ T = \frac{2\pi}{\omega} $. The correct expression for $ \omega $ is ( $ g $ is the acceleration due to gravity): 
Read More: Some Basic Concepts of Chemistry
There are two ways of classifying the matter:
Matter can exist in three physical states:
Based upon the composition, matter can be divided into two main types: