Q: 1 gram of sodium hydroxide was treated with 25 mL of 0.75 M HCl solution, the mass of sodium hydroxide left unreacted is equal to

aligned & M = W 1000 . M _2 V (in mL ) \\ & W = . M M _2 V (in mL ) 1000 = 0.75 36.5 25 1000 \\ &=0.684 ~g (Mass of HCl ) \\ & 36.5 ~g HCl + 40 ~g NaOH HCl + NaOH aligned 36.5 g HCl reacts with NaOH =40 ~g 0.684 g HCl reacts with NaOH = 40 36.5 0.684 0.750 ~g Amount of NaOH left =1 ~g -0.750 ~g =0.250 ~g =250 mg

Q: A compound X contains 32 \% of A, 20 \% of B and remaining percentage of C. Then, the empirical formula of X is : (Given atomic masses of A=64 ; B=40 ; C=32 u)

array |c|c|c|c|c| Element & array c Mass \\ percentage \% array & array c No. of \\ moles array & array c No. of moles / \\ Smallest number array & array c Simplest whole \\ number array \\ A & 32 \% & 32 64 = 1 2 & 1 2 2 & =1 \\ B & 20 \% & 20 40 = 1 2 & 1 2 2 & =1 \\ C & 48 \% & 48 32 = 3 2 & 3 2 2 & =3 \\ array So, empirical formula of X= array clll A & : & B & C \\ 1 & : & 1 & : array The correct empirical formula of compound X is ABC _3

Question 1

The numbers 17.0145 and 21.0235 were rounded to three figures after the decimal point. The resulting numbers, respectively, are

Accepted Answer

According to the rules for rounding off numbers, if the digit to be dropped is 5 and it is not followed by any non-zero digit, the preceding digit is left unchanged if it is even, and increased by 1 if it is odd. For 17.0145, the digit to be dropped is 5. The preceding digit is 4, which is even. Therefore, it remains unchanged. The rounded number is 17.014. For 21.0235, the digit to be dropped is 5. The preceding digit is 3, which is odd. Therefore, it is increased by 1. The rounded number is 21

Question 2

When 1 dm^3 of CO _2 gas is passed over hot coke, the volume of gaseous mixture after complete reaction at STP becomes 1.4 dm^3. The composition of the gaseous mixture at STP is :

Accepted Answer

The reaction of carbon dioxide with hot coke is given by: CO _2(g) + C (s) 2 CO (g) Initial volume of CO _2 = 1 dm^3 Let x dm^3 of CO _2 react with hot coke. Volume of CO _2 remaining = 1 - x dm^3 Volume of CO formed = 2x dm^3 Total volume of the gaseous mixture = (1 - x) + 2x = 1 + x dm^3 Given that the total volume after the reaction is 1.4 dm^3: 1 + x = 1.4 x = 0.4 dm^3 Volume of CO _2 remaining = 1 - 0.4 = 0.6 dm^3 Volume of CO formed = 2 0.4 = 0.8 dm^3 The composition of the gaseous mixture

Question 3

The number of hydrogen atoms present in 5.4 g of urea is : (Given : Molar mass of urea : 60 g mol^ -1 , N_A : 6.022 10^ 23 particles mol^ -1 )

Accepted Answer

The chemical formula of urea is NH_2CONH_2. Number of moles of urea = 5.4 60 = 0.09 mol. One molecule of urea contains 4 hydrogen atoms. Number of moles of hydrogen atoms = 4 0.09 = 0.36 mol. Total number of hydrogen atoms = 0.36 N_A = 0.36 6.022 10^ 23 = 2.16792 10^ 23 2.168 10^ 23 . Answer: 2.168 10^ 23

Question 4

The amount of carbon dioxide evolved upon complete combustion of 116 g of n-butane is (Given: atomic mass in amu H =1, C =12 and O =16)

Accepted Answer

The balanced chemical equation for the combustion of n-butane is: C_4H_ 10 + 13 2 O_2 4CO_2 + 5H_2O Molar mass of n-butane (C_4H_ 10 ) = 4 12 + 10 1 = 58 g/mol Number of moles of n-butane = 116 58 = 2 moles From the balanced equation, 1 mole of n-butane produces 4 moles of CO_2. Number of moles of CO_2 produced = 2 4 = 8 moles Molar mass of CO_2 = 12 + 2 16 = 44 g/mol Mass of CO_2 produced = 8 44 = 352 g Answer: 352 g

Question 5

Among the following choose the ones with equal number of atoms. A. 212 g of Na _2 CO _3( ~s ) [molar mass =106 ~g ] B. 248 g of Na _2 O (s) [molar mass =62 ~g ] C. 240 g of NaOH ( s ) [molar mass =40

Accepted Answer

No. of atoms = Atomicity mole N _ A (A) No. of atoms =6 212 106 N _ A =12 ~N _ A (B) No. of atoms =3 248 62 N _ A =12 ~N _ A (C) No. of atoms =3 240 40 N _ A =18 ~N _ A (D) No. of atoms =2 12 2 N _ A =12 ~N _ A (E) No. of atoms =3 220 44 N_A=15 N_A

Question 6

Dalton's Atomic theory could not explain which of the following?

Accepted Answer

Dalton's atomic theory could not explain law of gaseous volume.

Question 7

1.0 g of H _2 has same number of molecules as in:

Accepted Answer

Number of moles of H _2= 1 2 =0.5 Number of molecules of H _2=0.5 ~N _ A . (1) Number of moles of N _2= 14 28 =0.5 Number of molecules of N _2=0.5 ~N _ A (2) Number of moles of H _2 O = 18 18 =1 Number of molecules of H _2 O =1 N _ A = N _ A (3) Number of moles of CO = 16 28 = 4 7 Number of molecules of CO = 4 7 ~N _ A (4) Number of moles of N _2= 28 28 =1 Number of molecules of N _2=1 N _ A = N _ A

Question 8

1 gram of sodium hydroxide was treated with 25 mL of 0.75 M HCl solution, the mass of sodium hydroxide left unreacted is equal to

Accepted Answer

aligned & M = W 1000 . M _2 V (in mL ) \ & W = . M M _2 V (in mL ) 1000 = 0.75 36.5 25 1000 \ &=0.684 ~g (Mass of HCl ) \ & 36.5 ~g HCl + 40 ~g NaOH HCl + NaOH aligned 36.5 g HCl reacts with NaOH =40 ~g 0.684 g HCl reacts with NaOH = 40 36.5 0.684 0.750 ~g Amount of NaOH left =1 ~g -0.750 ~g =0.250 ~g =250 mg

Question 9

A compound X contains 32 \% of A, 20 \% of B and remaining percentage of C. Then, the empirical formula of X is : (Given atomic masses of A=64 ; B=40 ; C=32 u)

Accepted Answer

array |c|c|c|c|c| Element & array c Mass \ percentage \% array & array c No. of \ moles array & array c No. of moles / \ Smallest number array & array c Simplest whole \ number array \ A & 32 \% & 32 64 = 1 2 & 1 2 2 & =1 \ B & 20 \% & 20 40 = 1 2 & 1 2 2 & =1 \ C & 48 \% & 48 32 = 3 2 & 3 2 2 & =3 \ array So, empirical formula of X= array clll A & : & B & C \ 1 & : & 1 & : array The correct empirical formula of compound X is ABC _3

Question 10

The highest number of helium atoms is in

Accepted Answer

(1) 4 u of He = 4 u 4 u =1 He atom (2) 4 g of Helium = 4 ~g 4 ~g mole =1 mole = N _ A He atom (3) aligned 2.2710982 of He at STP & = 2.271 22.710982 mole \ & =0.1 ~mole \ & =0.1 ~N _ A He atom aligned (4) 4 mol of He =4 ~N _ A He atoms

Some Basic Concepts of Chemistry — Previous Year Questions & Solutions