IMO Practice Test: Some Basic Concepts of Chemistry | Class 11 Chemistry

Question 1 of 14

The number of oxygen atoms in $4.4\,g$ of $CO_2$ is:

$6.022\times10^{22}$

$1.204\times10^{23}$

$3.011\times10^{23}$

$6.022\times10^{23}$

Explanation: $n = 4.4/44 = 0.1\,mol$; molecules $= 6.022\times10^{22}$; each has 2 O atoms $\Rightarrow 1.204\times10^{23}$.

Question 2 of 14

$2.8\,g$ of $N_2$ reacts with excess $H_2$. The maximum mass of $NH_3$ formed ($N_2 + 3H_2 \rightarrow 2NH_3$) is:

$3.4\,g$

$1.7\,g$

$5.1\,g$

$17\,g$

Explanation: $n(N_2) = 2.8/28 = 0.1\,mol \Rightarrow 0.2\,mol\,NH_3 = 0.2\times17 = 3.4\,g$.

Question 3 of 14

Equal masses of $O_2$ and $SO_2$ are taken. The ratio of their number of molecules is:

$1:2$

$1:1$

$2:1$

$4:1$

Explanation: Molecules $\propto 1/M$. $M(O_2)=32$, $M(SO_2)=64$, so ratio $= 64:32 = 2:1$.

Question 4 of 14

$0.5\,mol$ of a metal carbonate $MCO_3$ weighs $50\,g$. The atomic mass of the metal $M$ is:

$24$

$56$

$100$

$40$

Explanation: $M(MCO_3) = 50/0.5 = 100$; subtract $CO_3 = 60 \Rightarrow M = 40$.

Question 5 of 14

A $0.2\,M$ solution is prepared from $0.05\,mol$ solute. The volume of solution is:

$250\,mL$

$100\,mL$

$500\,mL$

$1000\,mL$

Explanation: $V = n/M = 0.05/0.2 = 0.25\,L = 250\,mL$.

Question 6 of 14

The mass of carbon in $0.25\,mol$ of $C_6H_{12}O_6$ is:

$9\,g$

$12\,g$

$18\,g$

$36\,g$

Explanation: Carbon per mole $= 6\times12 = 72\,g$; for $0.25\,mol$, $= 0.25\times72 = 18\,g$.

Question 7 of 14

$8\,g$ of $O_2$ and $1\,g$ of $H_2$ are mixed for $2H_2 + O_2 \rightarrow 2H_2O$. The limiting reagent is:

$O_2$

$H_2$

neither (exact ratio)

water

Explanation: $n(O_2)=8/32=0.25\,mol$, $n(H_2)=1/2=0.5\,mol$. The required ratio $H_2:O_2$ is $2:1$, and $0.5:0.25 = 2:1$ exactly, so neither reactant is in excess.

Question 8 of 14

If $1\,L$ of a gas at STP weighs $1.25\,g$, its molar mass is approximately:

$22.4\,g\,mol^{-1}$

$32\,g\,mol^{-1}$

$28\,g\,mol^{-1}$

$44\,g\,mol^{-1}$

Explanation: $M = 1.25\times22.4 = 28\,g\,mol^{-1}$ (e.g. $N_2$ or CO).

Question 9 of 14

The molality of a solution that is $0.5\,M$ (density $1.0\,g\,mL^{-1}$, solute molar mass small, treat solvent mass ~ solution mass) is approximately:

$0.25\,m$

$1\,m$

$2\,m$

$0.5\,m$

Explanation: With density $1\,g\,mL^{-1}$ and dilute solute, $1\,L \approx 1\,kg$ solvent, so molality $\approx$ molarity $= 0.5\,m$.

Question 10 of 14

The number of electrons in $1.8\,g$ of water ($H_2O$ has 10 electrons per molecule) is:

$6.022\times10^{23}$

$6.022\times10^{22}$

$3.011\times10^{23}$

$1.204\times10^{23}$

Explanation: $n = 1.8/18 = 0.1\,mol$; molecules $= 6.022\times10^{22}$; electrons $= 10\times$ that $= 6.022\times10^{23}$.

Question 11 of 14

Two compounds of S and O contain S:O mass ratios of $1:1$ ($SO_2$) and $1:1.5$ ($SO_3$). This illustrates the law of:

multiple proportions

definite proportions

conservation of mass

combining volumes

Explanation: Different whole-number oxygen ratios for the same sulfur mass illustrate the law of multiple proportions.

Question 12 of 14

The mass of solute needed to make $250\,mL$ of $0.1\,M$ $NaOH$ is:

$0.4\,g$

$1.0\,g$

$2.0\,g$

$4.0\,g$

Explanation: Moles $= 0.1\times0.25 = 0.025\,mol$; mass $= 0.025\times40 = 1.0\,g$.

Question 13 of 14

The percentage by mass of water of crystallisation in $CuSO_4\cdot5H_2O$ (M = 250) is:

$18\%$

$45\%$

$36\%$

$64\%$

Explanation: Mass of $5H_2O = 90$; $\%\,= (90/250)\times100 = 36\%$.

Question 14 of 14

If $0.44\,g$ of $CO_2$ is produced, the number of moles of carbon burnt is:

$0.005$

$0.01$

$0.02$

$0.1$

Explanation: $n(CO_2) = 0.44/44 = 0.01\,mol$; one C per $CO_2 \Rightarrow 0.01\,mol$ carbon.

IMO Practice Test — Some Basic Concepts of Chemistry