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Chemistry NEET Questions — Free MCQ Practice

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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)

A. 20% and 80% B. 80% and 20% C. 25% and 75% D. 75% and 25%

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Explanation

Let the % of isotope with atomic mass 10.01 be â€˜xâ€™ % of isotope with atomic mass 11.01 = 100-x Avg at mass = $ { 10.01x + (100 - x)11.01 \over 100 } = 10.81(Given) $

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Calculate the mass percent of Na and S in sodium sulphate.

A. Na = 16.2%, S = 22.54% B. Na = 32.39%, S = 11.26% C. Na = 22.54%, S = 32.39% D. Na = 32.39%, S = 22.54%

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Explanation

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%.

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Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% oxygen by mass.

A. FeO B. $ Fe_2 O_3 $ C. $ Fe_2 O_4 $ D. $ Fe_3 O_2 $

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Explanation

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 â‰ˆ 2:3. Therefore, the empirical formula is Fe2O3.

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

A.

Half mole of N2

B.

Half mole of N2O

C.

1 mole of NO2

D.

1 mole of NH3

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Calculate the number of sulphate ions in 100mL of 0.001M ammonium sulphate solution.

A. $ 6.022 \times 10 ^{-19} $ B. $ 6.022 \times 10 ^{19} $ C. $ 6.022 \times 10 ^{20} $ D. $ 6.022 \times 10 ^{-20} $

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Explanation

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

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

A.

2.31 M

B.

0.213 M

C.

0.0213 M

D.

23.1 M

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Explanation

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

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The normality of 0.3M phosphorous acid is (IITJEE 1999)

A. 0.1 B. 0.9 C. 0.3 D. 0.6

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Explanation

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

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An aqueous solution of 6.3g oxalic acid dihydrate is made upto 250 mL. The volume of 0.1 N NaOH required to completely neutralize 10 mL of this solution is

A. 40 mL B. 20 mL C. 10 mL D. 4 mL

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Explanation

$ 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 $

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The pair of the compounds in which both the metals are in the highest possible oxidation state is

A. $ [ Fe ( CN)_6 ] ^ {3-} , [ Co ( CN)_6 ] ^ {3-} ] $ B. $ CrO_2 Cl_2 , MnO_4 ^ - $ C. $ TiO_3 , MnO_2 $ D. $ [ Co (CN)_6 ] ^ {3-} , MnO_3 $

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Explanation

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

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In the analysis of 0.0500 g sample of feldspar, a mixture of the chiorides of sodium and potassium is obtained, which weighs 0.1180 g. Subsequent treatment of the mixed chlorides with silver nitrate gives 0.2451g of silver chloride. What is the percentange of a sodium oxide and potassium oxide in feldspar ?

A. $ 10.62 \% Na_2O , 3.58 \% K_2O $ B. $ 3.58 \% Na_2O , 10.62 \% K_2O $ C. $ 10.62 \% Na_2O , 35.8 \% K_2O $ D. $ 35.8 \% Na_2O , 10.62 \% K_2O $

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Explanation

Suppose amount of NaCl in the mixture = â€˜xâ€™ 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 \% ....$

Chemistry MCQs for NEET — Practice Questions with Answers

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