The concentration of hydrogen ion in a sample of soft drink is
Question: The concentration of hydrogen ion in a sample of soft drink is $3.8 \times 10^{-3} \mathrm{M}$. what is its $\mathrm{pH}$ ? Solution: Given, $\left[\mathrm{H}^{+}\right]=3.8 \times 10^{-3} \mathrm{M}$ pH value of soft drink $=-\log \left[\mathrm{H}^{+}\right]$ $=-\log \left(3.8 \times 10^{-3}\right)$ $=-\log 3.8-\log 10^{-3}$ $=-\log 3.8+3$ $=-0.58+3$ $=2.42$...
Read More →Two dice, one blue and one grey, are thrown at the same time.
Question: Two dice, one blue and one grey, are thrown at the same time. (i) Write down all the possible outcomes and complete the following table: (ii) A student argues that 'there are 11 possible outcomes $2,3,4,5,6,7,8,9,10,11$ and 12 . Therefore, each of them has a probability $\frac{1}{11}$. Do you agree with this argument? Solution: (i) It can be observed that, To get the sum as 2, possible outcomes = (1, 1) To get the sum as 3, possible outcomes = (2, 1) and (1, 2) To get the sum as 4, pos...
Read More →Find the general solution of the equation $cos 4 x=cos 2 x$
Question: Find the general solution of the equation $\cos 4 x=\cos 2 x$ Solution: $\cos 4 x=\cos 2 x$ $\Rightarrow \cos 4 x-\cos 2 x=0$ $\Rightarrow-2 \sin \left(\frac{4 x+2 x}{2}\right) \sin \left(\frac{4 x-2 x}{2}\right)=0$ $\left[\because \cos \mathrm{A}-\cos \mathrm{B}=-2 \sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\right]$ $\Rightarrow \sin 3 x \sin x=0$ $\Rightarrow \sin 3 x=0 \quad$ or $\quad \sin x=0$ $\therefore 3 \mathrm{x}=\mathrm...
Read More →Show that the square of an odd positive integer is of the form 8q + 1, for some integer q.
Question: Show that the square of an odd positive integer is of the form 8q+ 1, for some integerq. Solution: To Prove: that the square of an odd positive integer is of the form 8q+ 1, for some integerq. Proof: Since any positive integernis of the form 4m+ 1 and 4m+ 3 Ifn= 4m+ 1 $\Rightarrow n^{2}=(4 m+1)^{2}$ $\Rightarrow n^{2}=(4 m)^{2}+8 m+1$ $\Rightarrow n^{2}=16 m^{2}+8 m+1$ $\Rightarrow n^{2}=8 m(2 m+1)+1$ $\Rightarrow n^{2}+8 q+1(q=m(2 m+1))$ Ifn= 4m+ 3 $\Rightarrow n^{2}=(4 m+3)^{2}$ $\Ri...
Read More →Consider the binary operation ∨ on the set {1, 2, 3, 4, 5} defined by a ∨b = min {a, b}.
Question: Consider the binary operation $\vee$ on the set $\{1,2,3,4,5\}$ defined by $a \vee b=\min \{a, b\}$. Write the operation table of the operation $v$. Solution: The binary operation on the set {1, 2, 3, 4, 5} is defined asab= min {a,b} mnForE;a,b {1, 2, 3, 4, 5}. Thus, the operation table for the given operation can be given as:...
Read More →Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base
Question: Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base: (a) $\mathrm{OH}^{-}$(b) $\mathrm{F}^{-}$(c) $\mathrm{H}^{+}$(d) $\mathrm{BCl}_{3}$. Solution: (a) $\mathrm{OH}^{-}$is a Lewis base since it can donate its lone pair of electrons. (b) $\mathrm{F}^{-}$is a Lewis base since it can donate a pair of electrons. (c) $\mathrm{H}^{+}$is a Lewis acid since it can accept a pair of electrons. (d) $\mathrm{BCl}_{3}$ is a Lewis acid since it c...
Read More →A lot consists of 144 ball pens of which 20 are defective and the others are good.
Question: A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that (i) She will buy it? (ii) She will not buy it? Solution: Total number of pens = 144 Total number of defective pens = 20 Total number of good pens = 144 20 = 124 (i) Probability of getting a good pen $=\frac{124}{144}=\frac{31}{36}$ $P($ Nuri buys...
Read More →Find the general solution of cosec x = –2
Question: Find the general solution of $\operatorname{cosec} x=-2$ Solution: $\operatorname{cosec} x=-2$ It is known that $\operatorname{cosec} \frac{\pi}{6}=2$ $\therefore \operatorname{cosec}\left(\pi+\frac{\pi}{6}\right)=-\operatorname{cosec} \frac{\pi}{6}=-2$ and $\operatorname{cosec}\left(2 \pi-\frac{\pi}{6}\right)=-\operatorname{cosec} \frac{\pi}{6}=-2$ i.e., $\operatorname{cosec} \frac{7 \pi}{6}=-2$ and $\operatorname{cosec} \frac{11 \pi}{6}=-2$ Therefore, the principal solutions are $x=\...
Read More →The species: H2O,, and NH3 can act both as Brönsted acids and bases.
Question: The species: $\mathrm{H}_{2} \mathrm{O}, \mathrm{HCO}_{7}^{-}, \mathrm{HSO}_{4}^{-}$, and $\mathrm{NH}_{3}$ can act both as Brnsted acids and bases. For each case give the corresponding conjugate acid and base. Solution: The table below lists the conjugate acids and conjugate bases for the given species....
Read More →Suppose you drop a die at random on the rectangular region shown in the given figure.
Question: Suppose you drop a die at random on the rectangular region shown in the given figure. What is the probability that it will land inside the circle with diameter 1 m? Solution: Area of rectangle $=1 \times b=3 \times 2=6 \mathrm{~m}^{2}$ Area of circle (of diameter $1 \mathrm{~m})=\pi r^{2}=\pi\left(\frac{1}{2}\right)^{2}=\frac{\pi}{4} \mathrm{~m}^{2}$ $P($ die will land inside the circle $)=\frac{\frac{\pi}{4}}{6}=\frac{\pi}{24}$...
Read More →Write the conjugate acids for the following Brönsted bases:
Question: Write the conjugate acids for the following Brnsted bases: $\mathrm{NH}_{2}^{-}, \mathrm{NH}_{3}$ and $\mathrm{HCOO}^{-}$. Solution: The table below lists the conjugate acids for the given Bronsted bases....
Read More →A child has a die whose six faces shows the letters as given below:
Question: A child has a die whose six faces shows the letters as given below: The die is thrown once. What is the probability of getting (i) A? (ii) D? Solution: Total number of possible outcomes on the dice = 6 (i) Total number of faces having A on it = 2 $P($ getting $A)=\frac{2}{6}=\frac{1}{3}$ (ii) Total number of faces having $D$ on it $=1$ $P($ getting $D)=\frac{1}{6}$...
Read More →Prove that the square of any positive integer is of the form 5q, 5q + 1, 5q + 4 for some integer q.
Question: Prove that the square of any positive integer is of the form 5q, 5q+ 1, 5q+ 4 for some integerq. Solution: To Prove: that the square of any positive integer is of the form 5qor 5q+ 1, 5q+ 4 for some integerq. Proof: Since positive integernis of the form of 5qor 5q+ 1, 5q+ 4 Ifn= 5q Then, $n^{2}=(5 q)^{2}$ $\Rightarrow \quad n^{2}=25 q^{2}$ $\Rightarrow \quad n^{2}=5(5 q)$ $\Rightarrow \quad n^{2}=5 m($ where $m=5 q)$ Ifn= 5q+ 1 Then, $n^{2}=(5 q+1)^{2}$ $\Rightarrow \quad n^{2}=(5 q)^{...
Read More →What will be the conjugate bases for the Brönsted acids
Question: What will be the conjugate bases for the Brnsted acids: $\mathrm{HF}, \mathrm{H}_{2} \mathrm{SO}_{4}$ and $\mathrm{HCO}_{3}$ ? Solution: The table below lists the conjugate bases for the given Bronsted acids....
Read More →A box contains 90 discs which are numbered from 1 to 90.
Question: A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number (ii) a perfect square number (iii) a number divisible by 5. Solution: Total number of discs = 90 (i) Total number of two-digit numbers between 1 and 90 = 81 $P$ (getting a two-digit number) $=\frac{81}{90}=\frac{9}{10}$ (ii) Perfect squares between 1 and 90 are 1, 4, 9, 16, 25, 36, 49, 64, and 81. Therefore, total number of perf...
Read More →Which of the followings are Lewis acids?
Question: Which of the followings are Lewis acids? $\mathrm{H}_{2} \mathrm{O}, \mathrm{BF}_{3}, \mathrm{H}^{+}$, and $\mathrm{NH}_{4}^{+}$ Solution: Lewis acids are those acids which can accept a pair of electrons. For example, $\mathrm{BF}_{3}, \mathrm{H}^{+}$, and $\mathrm{NH}_{4}^{+}$are Lewis acids....
Read More →A lot of 20 bulbs contain 4 defective ones.
Question: (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective? (ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective? Solution: (i) Total number of bulbs = 20 Total number of defective bulbs = 4 $P($ getting a defective bulb $)=\frac{\text { Number of favourable outcomes }}{\text { Numhe...
Read More →What is meant by the conjugate acid-base pair?
Question: What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species: $\mathrm{HNO}_{2}, \mathrm{CN}^{-}, \mathrm{HClO}_{4}, \mathrm{~F}^{-}, \mathrm{OH}^{-}, \mathrm{CO}_{3}^{2-}$, and $\mathrm{S}^{-}$ Solution: A conjugate acid-base pair is a pair that differs only by one proton. The conjugate acid-base for the given species is mentioned in the table below....
Read More →Find the principal and general solutions of the equation $cot x=-sqrt{3}$
Question: Find the principal and general solutions of the equation $\cot x=-\sqrt{3}$ Solution: $\cot x=-\sqrt{3}$ It is known that $\cot \frac{\pi}{6}=\sqrt{3}$ $\therefore \cot \left(\pi-\frac{\pi}{6}\right)=-\cot \frac{\pi}{6}=-\sqrt{3}$ and $\cot \left(2 \pi-\frac{\pi}{6}\right)=-\cot \frac{\pi}{6}=-\sqrt{3}$ i.e., $\cot \frac{5 \pi}{6}=-\sqrt{3}$ and $\cot \frac{11 \pi}{6}=-\sqrt{3}$ Therefore, the principal solutions are $x=\frac{5 \pi}{6}$ and $\frac{11 \pi}{6}$. Now, $\cot x=\cot \frac{5...
Read More →For each binary operation * defined below, determine whether * is commutative or associative.
Question: For each binary operation * defined below, determine whether * is commutative or associative. (i) On $\mathbf{Z}$, define $a^{*} b=a-b$ (ii) On Q, define $a^{*} b=a b+1$ (iii) On $\mathbf{Q}$, define $a^{*} b=\frac{a b}{2}$ (iv) On Z^ , define $a^{*} b=2^{a b}$ (v) On $\mathbf{Z}^{+}$, define $a^{*} b=a^{b}$ (vi) On $\mathbf{R}-\{-1\}$, define $a * b=\frac{a}{b+1}$ Solution: (i) On $\mathbf{Z},{ }^{*}$ is defined by $a{ }^{*} b=a-b$. It can be observed that $1^{*} 2=1-2=1$ and $2^{*} 1...
Read More →12 defective pens are accidentally mixed with 132 good ones.
Question: 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one. Solution: Total number of pens = 12 + 132 = 144 Total number of good pens = 132 $P$ (getting a good pen) $=\frac{\text { Number of favourable outcomes }}{\text { Number of total possible outcomes }}$ $=\frac{132}{144}=\frac{11}{12}$...
Read More →The reaction, CO(g) + 3H2(g)CH4(g) + H2O(g) is at equilibrium at 1300 K in a 1L flask.
Question: The reaction, $\mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longleftrightarrow \mathrm{CH}_{4}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ is at equilibrium at $1300 \mathrm{~K}$ in a $1 \mathrm{~L}$ flask. It also contain $0.30 \mathrm{~mol}$ of $\mathrm{CO}, 0.10 \mathrm{~mol}$ of $\mathrm{H}_{2}$ and $0.02 \mathrm{~mol}$ of $\mathrm{H}_{2} \mathrm{O}$ and an unknown amount of $\mathrm{CH}_{4}$ in the flask. Determine the concentration of $\mathrm{CH}_{4}$ in the mi...
Read More →Five cards−−the ten, jack, queen, king and ace of diamonds,
Question: Five cardsthe ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random. (i) What is the probability that the card is the queen? (ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen? Solution: (i) Total number of cards = 5 Total number of queens = 1 $P($ getting a queen $)=\frac{\text { Number of favourable outcomes }}{\text { Number of total possib...
Read More →Find the principal and general solutions of the equation $sec x=2$
Question: Find the principal and general solutions of the equation $\sec x=2$ Solution: $\sec x=2$ It is known that $\sec \frac{\pi}{3}=2$ and $\sec \frac{5 \pi}{3}=\sec \left(2 \pi-\frac{\pi}{3}\right)=\sec \frac{\pi}{3}=2$ Therefore, the principal solutions are $x=\frac{\pi}{3}$ and $\frac{5 \pi}{3}$ Now, $\sec x=\sec \frac{\pi}{3}$ $\Rightarrow \cos x=\cos \frac{\pi}{3} \quad\left[\sec x=\frac{1}{\cos x}\right]$ $\Rightarrow \mathrm{x}=2 \mathrm{n} \pi \pm \frac{\pi}{3}$, where $\mathrm{n} \i...
Read More →Prove that the square of any positive integer is of the form 4q or 4q + 1 for some integer q.
Question: Prove that the square of any positive integer is of the form 4qor 4q+1 for some integerq. Solution: To Prove: that the square of any positive integer is of the form 4qor 4q+ 1 for some integerq. Proof: Since positive integernis of the form of 2qor 2q+ 1 Ifn= 2q Then, $n^{2}=(2 q)^{2}$ $\Rightarrow \quad n^{2}=4 q^{2}$ $\Rightarrow \quad n^{2}=4 m\left(\right.$ where $\left.m=q^{2}\right)$ Ifn= 2q+ 1 Then, $n^{2}=(2 q+1)^{2}$ $\Rightarrow n^{2}=(2 q)^{2}+4 q+1$ $\Rightarrow n^{2}=4 q^{2...
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