Some Basic Concepts of ChemistryNEET Chemistry · Class 11 · NCERT Chapter 1

Solution

The reaction is ${\text{H}}_2+{\text{Cl}}_2\to 2\text{HCl}$. At STP, 22.4 L of ${\text{H}}_2$ is 1 mole and 11.2 L of ${\text{Cl}}_2$ is 0.5 mole. Since ${\text{Cl}}_2$ is the limiting reagent, 0.5 mole of ${\text{Cl}}_2$ will produce 1 mole of $\text{HCl}$, so option (a) is correct.

Solution

The balanced equation is $2\text{Mg}+{\text{O}}_2\to 2\text{MgO}$. Using stoichiometry, $1.0\text{g Mg}\Rightarrow \frac{1.0}{24}\times 32=1.33{\text{g O}}_2$ needed, but only $0.56{\text{g O}}_2$ is available. Thus, ${\text{O}}_2$ is the limiting reagent. The amount of $\text{Mg}$ used is $\frac{0.56}{32}\times 24=0.42\text{g}$. Therefore, $\text{Mg}$ left = $1.0-0.42=0.58\text{g}$. However, the closest option is (a) $\text{Mg},0.16\text{g}$, which is incorrect. The correct answer should be $\text{Mg},0.58\text{g}$, but the closest option is (a) $\text{Mg},0.16\text{g}$.

Solution

In the Kjeldahl's method, the reaction is ${\text{2NH}}_3+{\text{H}}_2{\text{SO}}_4\to ({\text{NH}}_4{)}_2{\text{SO}}_4$. The number of moles of ${\text{H}}_2{\text{SO}}_4$ used is $0.01\times 1=0.01$ moles. Since 2 moles of ${\text{NH}}_3$ neutralize 1 mole of ${\text{H}}_2{\text{SO}}_4$, the moles of ${\text{NH}}_3$ are $0.01\times 2=0.02$ moles. The mass of nitrogen in ${\text{NH}}_3$ is $0.02\times 14=0.28$ g. The percentage of nitrogen in the sample is $\frac{0.28}{0.75}\times 100=37.33\%$, so option (a) is correct.

Solution

∵ 1 mole water =6.02×1023 moleucles ∴ 18 mole water =18×6.02×1023 molecules So, 18 mole water has maximum number of molecules. www.vedantu.com 64

Solution

∵ Mass of 1 mol (6.022×1023 atoms) of carbon = 12g If Avogadro Number (NA) is changed then mass of 1 mol (6.022×1020 atom) of carbon =12×6.022×1020 6.022×1023 =12×10−3 g

Solution

Phosphinic acid is Hypophosphorous acid 𝐻3𝑃𝑂2 which is Monobasic acid. Phosphonic acid is phosphorous acid 𝐻3𝑃𝑂3 which is Dibasic acid.

Solution

– Nucleic acids are polymers of nucleotides – Proteins are polymers of amino acids – Polysaccharides are polymers of monosaccharides – Lipids are the esters of fatty acids and alcohol www.vedantu.com 22

Solution

(a) Cu+ goes into Cu+2 and Cu. (b) Mn+6 goes into Mn+7 and Mn+4

Solution

- ${\text{CO(g) + H}}_2\text{(g)}$ forms synthesis gas (iii).
- Temporary hardness is due to ${\text{Mg(HCO}}_3{\text{)}}_2$ and ${\text{Ca(HCO}}_3{\text{)}}_2$ (i).
- ${\text{B}}_2{\text{H}}_6$ is an electron-deficient hydride (ii).
- ${\text{H}}_2{\text{O}}_2$ has a non-planar structure (iv).

Solution

Unununnium is the temporary name for the element with atomic number 111, which has the official name Darmstadtium. The other matches are correct: Unnilunium (Mendelevium), Unniltrium (Lawrencium), and Unnilhexium (Seaborgium). Therefore, option (d) is incorrect.

Solution

Passing HCl through a solution of ${\text{CaCl}}_2$, ${\text{MgCl}}_2$, and $\text{NaCl}$ causes the formation of insoluble chlorides of ${\text{Ca}}^{2+}$ and ${\text{Mg}}^{2+}$, which crystallize out. $\text{NaCl}$ remains soluble. Thus, option (a) is correct.

Solution

To find the number of atoms, use $n=\frac{m}{M}\cdot N_A$, where $m$ is mass, $M$ is molar mass, and $N_A$ is Avogadro's number. For 1 g of each:
- Ag: $\frac{1}{108}\cdot N_A$
- Mg: $\frac{1}{24}\cdot N_A$
- O${}_2$: $\frac{1}{32}\cdot N_A$ (since O${}_2$ has a molar mass of 32 g/mol)
- Li: $\frac{1}{7}\cdot N_A$

Li has the highest number of atoms, so option (d) is correct.

Solution

Kjeldahl method is not applicable to compounds containing nitrogen in nitro group, azo groups and nitrogen present in the ring (e.g., pyridine) as nitrogen of these compounds does not change to ammonium sulphate under these conditions.

Solution

Let m gram mass of CaCO3 is required Pure CaCO3 in m gram = 95 m100 × Moles of CaCO3 = 95 m 100 100× Moles of HCl required = 2 × moles of CaCO3 = 95 m2 100 100×× 95 m 502 0.5100 100 1000× × = × m = 1.315 g ≈ 1.32 g

Solution

Option (3) is correct because maltose is a disaccharide formed by dehydration process i.e. , synthesis by elimination of one water molecule to form a glycosidic bond in between two glucose molecules. So, its molecular formula is. 26 12 6 12 22 11HOC HO 2 CHO×   → - 59 - NEET (UG)-2022 (Code- Q1)

Solution

Option (4) is the correct answer as glycogen is a polysaccharide and is a storage product in animals. • Globulins form antibodies which are also known as immunoglobulins. • Steroids form hormones like testosterone. • Thrombin is a biocatalyst which converts soluble fibrinogen to insoluble fibrin. - 65 - NEET (UG)-2022 (Code- Q1)

Solution

The reaction ${\text{CaCO}}_3\to \text{CaO}+{\text{CO}}_2$ shows that 1 mole of ${\text{CaCO}}_3$ produces 1 mole of ${\text{CO}}_2$. The molar mass of ${\text{CaCO}}_3$ is $40+12+3\times 16=100\text{g/mol}$. Given 20 g of 20% pure limestone, the mass of ${\text{CaCO}}_3$ is $20\times 0.20=4\text{g}$. The moles of ${\text{CaCO}}_3$ are $\frac{4}{100}=0.04\text{mol}$. Thus, the moles of ${\text{CO}}_2$ produced are also 0.04 mol. The molar mass of ${\text{CO}}_2$ is $12+2\times 16=44\text{g/mol}$. Therefore, the mass of ${\text{CO}}_2$ is $0.04\times 44=1.76\text{g}$, so option (c) is correct.

Solution

(1) In ozone; there are two resonating structures. (2) BF3 i.e., ; Dipole moment = 0 (3) (4)

Solution

Element Mass percentage % No. of moles No. of moles/ Smallest number Simplest whole number A 32% 32 1 64 2= 1 22 × = 1 B 20% 20 1 40 2= 1 22 × = 1 C 48% 48 3 32 2= 3 22 × = 3 So, empirical formula of A : B : CX 1 : 1 : 3= ∴ The correct empirical formula of compound X is ABC3 w BOTANY SECTION-A

Solution

Calculate the number of moles for each substance to determine the number of atoms.
- A: $212\text{g}{\text{Na}}_2{\text{CO}}_3/106\text{g/mol}=2\text{mol}\Rightarrow 2\times 6.022\times 10^{23}\times 6=7.2264\times 10^{24}$ atoms.
- B: $248\text{g}{\text{Na}}_2\text{O}/62\text{g/mol}=4\text{mol}\Rightarrow 4\times 6.022\times 10^{23}\times 3=7.2264\times 10^{24}$ atoms.
- C: $240\text{g}\text{NaOH}/40\text{g/mol}=6\text{mol}\Rightarrow 6\times 6.022\times 10^{23}\times 4=1.44528\times 10^{25}$ atoms.
- D: $12\text{g}{\text{H}}_2/2\text{g/mol}=6\text{mol}\Rightarrow 6\times 6.022\times 10^{23}\times 2=7.2264\times 10^{24}$ atoms.
- E: $220\text{g}{\text{CO}}_2/44\text{g/mol}=5\text{mol}\Rightarrow 5\times 6.022\times 10^{23}\times 3=9.033\times 10^{24}$ atoms.
A, B, and D have the same number of atoms, so option (a) is correct.

Solution

Compute the number of atoms (n × N_A × atoms-per-formula-unit) for each option and find pairs that match. Standard NEET trick: convert to moles, then multiply by atoms-per-molecule.

Per the NTA answer key the correct pair is the one where 'mol × atoms per formula unit' is identical for both options listed.