Practice free Some Basic Concepts in Chemistry (Chemistry) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
An organic compound gave the following results on analysis. C = 53.3%, H = 15.6%, N = 31.1%. Find molecular formula of compound. (Molecular Weight = 45)
An organic compound contains C, H and O in the proportion of 6 : 1 : 8 by weight, respectively. Find its molecular formula (V.D = 30)
Elements No. of moles Simple ratio C 6/12 = 0.5 1 H 1/1 = 1 2 O 8/16 = 0.5 1 $ \therefore E.F = CH_2O $ M.W = 2VD =60 $ \therefore n = 2 \therefore M.F = C_2H_4O_2 $
30 gm of Mg and 30 gm of $ O_2 $ are reacted and the residual mixure contains
In the reaction $ 2Al_{(s)}+ 6HCl_{(aq)} \rightarrow 2 Al ^{3+}_{(aq)} + 6Cl ^ - _{(aq)} + 3H_{2(g)} $ (AIEEE 2007)
Consider a titration of potassium dichromate solution with acidified Mohr’s salt solution using diphenyl amine as indicator. The number of moles of Mohr’s salt required per mole of dichromate is (IIT JEE 2007)
The redox reaction between potassium dichromate and Mohr’s salt is : $ 6Fe^{2+} + Cr_2 O^{2-} _7 + 14H ^ + \rightarrow 6Fe ^ {3+} 2Cr ^ {3+} + 7H_2 O $
Which has maximum number of atoms ? (IIT JEE 2003)
$ Number of particles \alpha Number of moles $ $ No. of moles of carbon = 24 \12 = 2 $
Number of atoms in 560 g of Fe (atomic mass = 56) is (AIEEE 2002)
In the standardization of $ Na_2S_2O_3 using K_2Cr_2O_7 by iodometry, the equivalent weight of K_2Cr_2O_7 $ is (IIT JEE 2001)
$ During the reaction, Cr_2O_7 ^{ 2-} changes to Cr^{3+} $. Hence the change in oxidation number of Cr is 6. $ \therefore Equivalent weight = { Molar mass \over 6 } $
Mixture X=0.02 mole of $ [Co(NH_3)_5SO_4] Br and 0.02 mole of[Co(NH_3)_5Br]SO_4 was prepared in 2L of Solution 1L of mixture X + excess AgNO_3 \rightarrow Y 1L of mixture X+ excess BaCl_2 \rightarrow Z $ Number of mole of Y and Z are (IIT JEE 2003)
$ In 2L solution, there are 0.02 mol Br ^ - ions and 0.02 mole so ^{ 2-} _4 $ $ \therefore 1 L of mixture X contains 0.01 mol Br ^ – and 0.01 mol SO ^{2-} _{4} $ ions. $ Hence, Y= 0.01 mol Ag Br Z= 0.01 mol BaSO_4 $
How many moles of electron weight one kilogram ? (IIT JEE 2002)
$ Mass of an electron = 9.108 \times 10 ^ {31} Kg $ $ No. of electron s in 1 Kg = { 1 \over 9.108 \times 10 ^ {-31} } $ $ = { 1 \over 9.108 \times 10 ^ {-31 } \times 6.023 \times 10^ {23} mol ^ {-1} }$ $ = { 10 ^ 8 \over 9.108 \times 6.023 } mol $
Ready to ace NEET?
Free access · No credit card required
Yes. You can attempt every Some Basic Concepts in Chemistry question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.
No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.
The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.