Some Basic Concepts in Chemistry MCQs for NEET — Chemistry Questions with Answers

The ratio by weight = 1 / 4 $ \therefore Ratio of moles = { 1/32 \over 4 /28 } = { 28 \over 4 \times 32 } = { 7 \over 32 } $

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).

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) $

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%.

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.

$$ No of moles of (NH_4)_2 SO_4 = molarity \times Vol (L) $$ $ = 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} $

$$ X_{ETOH} = { n_{(ETOH)} \over n_{ETOH} + n_{(H_2O)} $$ $$ \therefore 0.04 = { n_{(ETOH)} \over n_{(ETOH)} + 55.55 } $$ $$ \therefore n_{(ETOH)} = 2.31 $$

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