NEET Chemistry Questions: Some Basic Concepts in Chemistry

Pratcice NEET questions from all capters from huge question bank for free. All MCQs are based on NCERT syllabus. To practice from a specific subject and chapter, select a subject below. Login to practice in a structured way with explanations, bookmarks, lists, notes etc. Click here to Login or Sign up for free.

Physics Chemistry Botany Zoology

Question bank:

In chemistry a number is represented in the form $ N \times 10^n $ . This method of expressing the number is called scientific notation. What is the value of ‘N’ here.




No explanation available.
What is the correct scientific notation for 0.00016 ?




No explanation available.
How many significant digits are there in 0.25 ?




Significant figures are the digits that carry meaning and contribute to the value of a number. In 0.25, both digits (2 and 5) are significant figures, so the answer is 2 significant digits.

What is the number of neutrons in $ Zn^{2+} $ ion (Atomic mass number = 70) (IITJEE 1979)





No explanation available.
Find the total number of electrons in one molecule of carbon dioxide.




No explanation available.
A gaseous mixture contains oxygen and nitrogen in the ratio of 1 : 4 by weight. Therefore, the ratio of their number of molecules is




The ratio by weight = 1 / 4 $ \therefore Ratio of moles = { 1/32 \over 4 /28 } = { 28 \over 4 \times 32 } = { 7 \over 32 } $
Identify the incorrect unit conversion factor.




The correct unit conversion factors are: 1 cm^3 = 1 mL, 1 cm = 10 mm, and 60 s = 1 min. Since all the given options are incorrect, the answer is 'None of these'.

$ 90 g KClO_3 $ on heating gives 2.96g KCl and 1.92g oxygen. Which of the following laws is illustrated by this statement ?





The statement illustrates the law of mass conservation, which states that matter can neither be created nor destroyed in a chemical reaction. The total mass of reactants (90 g KClO3) is equal to the total mass of products (2.96 g KCl + 1.92 g O2).
Naturally occuring Boron consists of two isotopes having atomic masses 10.01 and 11.01 respectively. Calculate the percentage of both the isotopes in natural Boron (Atomic mass of natural Boron = 10.81)




Let the % of isotope with atomic mass 10.01 be ‘x’ % of isotope with atomic mass 11.01 = 100-x Avg at mass = $ { 10.01x + (100 - x)11.01 \over 100 } = 10.81(Given) $
Calculate the mass percent of Na and S in sodium sulphate.




To calculate the mass percent of Na and S in sodium sulphate (Na2SO4), we need to find the atomic masses of Na, S, and O and then calculate their percentages. The atomic masses are: Na = 23, S = 32, O = 16. The molar mass of Na2SO4 is (2 x 23) + 32 + (4 x 16) = 142. The mass percent of Na = (2 x 23)/142 x 100% = 32.39% and the mass percent of S = 32/142 x 100% = 22.54%.
Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% oxygen by mass.




To determine the empirical formula, we need to find the simplest whole number ratio of atoms. Given: 69.9% iron and 30.1% oxygen. Assume 100 g of the oxide. Iron = 69.9 g, Oxygen = 30.1 g. Divide by atomic masses: Fe = 69.9/55.85 = 1.25, O = 30.1/16 = 1.88. Ratio is 1.25:1.88 ≈ 2:3. Therefore, the empirical formula is Fe2O3.

In a reaction formula of electrons are transferred to one mole of HNO3 when it reacts as an oxidant.The possible reduction product is





No explanation available.
Calculate the number of sulphate ions in 100mL of 0.001M ammonium sulphate solution.




$$ No of moles of (NH_4)_2 SO_4 = molarity \times Vol (L) $$ $ = 0.001 \times 0.1 = 0.0001 $ $ \therefore No. of SO ^ {2-} _ 4 ions = 0.0001 \times 6.022 = 10 ^ {23} = 6.022 \times 10 ^ {19} $

Calculate the molarity of a solution of ethanol in water in which mole fraction of ethanol is 0.040.





$$ X_{ETOH} = { n_{(ETOH)} \over n_{ETOH} + n_{(H_2O)} $$ $$ \therefore 0.04 = { n_{(ETOH)} \over n_{(ETOH)} + 55.55 } $$ $$ \therefore n_{(ETOH)} = 2.31 $$
The normality of 0.3M phosphorous acid is (IITJEE 1999)




phosphorous acid $ (H_3PO_3 )$ is a dibasic acid. Its structure is as follows : $ Normality = basicity \times Molarity = 2 \times 0.3 = 0.6 $