NEET Practice Questions (MCQs) with Answers & Solutions

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 $

$ Equivalents of H_2C_2O_4 . 2H_2O in 10ml = Equivalents of NaOH $ $ \therefore { 6.3 \times 1, 0000 \over 63 \times 250 \times 0.1 } = V = 40mL $

The oxidation states of various metals are : (a) Fe = + 3, Co + 3 (b) Cr = + 6 , Mn + 7 (c) Ti = + 6, Mn + 4 (d) Co = + 3, Mn + 6

Suppose amount of NaCl in the mixture = ‘x’ g The amount of KCl in the mixture = (0.118 - x) g $ NaCl + AgNO_3 \rightarrow AgCl + NaNO_3 $ 58.5 143.5 $ \therefore x { 143.5 \times x \over 58.5 } g ..... (i) $ $ Similarly AgCl obtained from KCl = { 143.5 \times (0.118 - x) \over 74.5 } g... (ii)$ But (i) + (ii) = 0.2451 g (Given) Amount of NaCl = 0.0338 g Amount of KCl = 0.0842 g Now, $ 2NaCl =Na_2O $ 117 62 0.0338 $ {0.0338 \times 62 \over 117 } = 0.0179 g$ $ % of Na_2O = { 0.0179 \times 100 \over 0.5 } = 3.58 \% ....$

$ Equivalent weight = { Atomic Weight \over Valency } $

$ m = { 1000 M \over 1000d - MM_s } $ M = Molarity of solution d = density of solution $ M_s $ = Molar mass of solute $ = { 1000 \times 3 \over 1000 \times 1.25 - 3 \times 158 } = 3.865 $