Solve the following
Question: The solubility product of $\mathrm{Cr}(\mathrm{OH})_{3}$ at $298 \mathrm{~K}$ is $6.0 \times 10^{-31}$. The concentration of hydroxide ions in a saturated solution of $\mathrm{Cr}(\mathrm{OH})_{3}$ will be:$\left(2.22 \times 10^{-31}\right)^{1 / 4}$$\left(18 \times 10^{-31}\right)^{1 / 4}$$\left(18 \times 10^{-31}\right)^{1 / 2}$$\left(4.86 \times 10^{-29}\right)^{1 / 4}$Correct Option: , 2 Solution:...
Read More →Find the sum of all three-digit natural numbers which are divisible by 13.
Question: Find the sum of all three-digit natural numbers which are divisible by 13. Solution: All three-digit numbers which are divisible by 13are 104, 117, 130, 143,..., 988. This is an AP in whicha= 104,d= (117 - 104) = 13 andl= 988 Let the number of terms ben.Then Tn= 988⇒a+ (n- 1)d= 988⇒104 + (n-1)⨯ 13 = 988⇒ 13n =897⇒n=69 $\therefore$ Required sum $=\frac{n}{2}(a+l)$ $=\frac{69}{2}[104+988]=69 \times 546=37674$ Hence, the required sum is 37674....
Read More →Solve the following
Question: The $\mathrm{K}_{\mathrm{sp}}$ for the following dissociation is $1.6 \times 10^{-5}$ $\mathrm{PbCl}_{2(\mathrm{~s})} \rightleftharpoons \mathrm{Pb}_{(a q)}^{2+}+2 \mathrm{Cl}_{(a q)}^{-}$ Which of the following choices is correct for a mixture of $300 \mathrm{~mL} 0.134 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}$ and $100 \mathrm{~mL} 0.4 \mathrm{M} \mathrm{NaCl}$ ?Not enough data provided$\mathrm{Q}\mathrm{K}_{\mathrm{sp}}$$\mathrm{Q}\mathrm{K}_{\mathrm{sp}}$$\mathrm{Q}=\...
Read More →Find the sum of all multiples of 9 lying between 300 and 700.
Question: Find the sum of all multiples of 9 lying between 300 and 700. Solution: The multiples of 9 lying between 300 and 700 are 306, 315, ..., 693.This is an AP witha= 306,d= 9 andl= 693.Suppose there arenterms in the AP. Then, $a_{n}=693$ $\Rightarrow 306+(n-1) \times 9=693 \quad\left[a_{n}=a+(n-1) d\right]$ $\Rightarrow 9 n+297=693$ $\Rightarrow 9 n=693-297=396$ $\Rightarrow n=44$ $\therefore$ Required sum $=\frac{44}{2}(306+693) \quad\left[S_{n}=\frac{n}{2}(a+l)\right]$ $=22 \times 999$ $=...
Read More →Find the sum of first 15 multiples of 8.
Question: Find the sum of first 15 multiples of 8. Solution: The first 15 multiples of 8 are 8, 16, 24, 32,... This is an AP in whicha= 8,d= (16 - 8) = 8 andn= 15.Thus, we have: $l=a+(n-1) d$ $=8+(15-1) 8$ $=120$ $\therefore$ Required sum $=\frac{n}{2}(a+l)$ $=\frac{15}{2}[8+120]=15 \times 64=960$ Hence, the required sum is 960....
Read More →For the following Assertion and Reason,
Question: For the following Assertion and Reason, the correct option is: Assertion: The $\mathrm{pH}$ of water increases with increase in temperature. Reason: The dissociation of water into $\mathrm{H}^{+}$and $\mathrm{OH}^{-}$is an exothermic reaction.Both assertion and reason are true, and the reason is the correct explanation for the assertion.Both assertion and reason are false.Both assertion and reason are true, but the reason is not the correct explanation for the assertion.Assertion is no...
Read More →Find the sum of first forty positive integers divisible by 6.
Question: Find the sum of first forty positive integers divisible by 6. Solution: The positive integers divisible by 6 are 6, 12, 18, ... .This is an AP witha= 6andd= 6.Also,n= 40 (Given) Using the formula, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$, we get $S_{40}=\frac{40}{2}[2 \times 6+(40-1) \times 6]$ $=20(12+234)$ $=20 \times 246$ $=4920$ Hence, the required sum is 4920....
Read More →Solve the following
Question: $3 \mathrm{~g}$ of acetic acid is added to $250 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{HCI}$ and the solution made up to $500 \mathrm{~mL}$. To $20 \mathrm{~mL}$ of this solution $\frac{1}{2} \mathrm{~mL}$ of $5 \mathrm{M} \mathrm{NaOH}$ is added. The $\mathrm{pH}$ of the solution is___________________. [Given: pKa of acetic acid $=4.75$, molar mass of acetic acid $=60 \mathrm{~g} / \mathrm{mol}, \log 3=0.4771$ ] Neglect any changes in volume. Solution: (5.22)...
Read More →If the solubility product of
Question: If the solubility product of $\mathrm{AB}_{2}$ is $3.20 \times 10^{-11} \mathrm{M}^{3}$, then the solubility of $\mathrm{AB}_{2}$ in pure water is__________ $\times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$. [Assuming that neither kind of ion reacts with water] Solution:...
Read More →Find the sum of all natural numbers between 200 and 400 which are divisible by 7.
Question: Find the sum of all natural numbers between 200 and 400 which are divisible by 7. Solution: Natural numbers between 200 and 400 which are divisible by 7 are 203, 210, ..., 399.This is an AP witha= 203,d= 7 andl= 399.Suppose there arenterms in the AP. Then, $a_{n}=399$ $\Rightarrow 203+(n-1) \times 7=399 \quad\left[a_{n}=a+(n-1) d\right]$ $\Rightarrow 7 n+196=399$ $\Rightarrow 7 n=399-196=203$ $\Rightarrow n=29$ $\therefore$ Required sum $=\frac{29}{2}(203+399) \quad\left[S_{n}=\frac{n}...
Read More →A small spherical droplet of density d
Question: A small spherical droplet of density $d$ is floating exactly half immersed in a liquid of density $\rho$ and surface tension T. The radius of the droplet is (take note that the surface tension applies an upward force on the droplet):(1) $r=\sqrt{\frac{2 \mathrm{~T}}{3(d+\rho) g}}$(2) $r=\sqrt{\frac{\mathrm{T}}{(d-\rho) g}}$(3) $r=\sqrt{\frac{\mathrm{T}}{(d+\rho) g}}$(4) $r=\sqrt{\frac{3 \mathrm{~T}}{(2 d-\rho) g}}$Correct Option: , 4 Solution: (4) For the drops to be in equilibrium upw...
Read More →Arrange the following solutions in the decreasing order of pOH:
Question: Arrange the following solutions in the decreasing order of pOH: (A) $0.01 \mathrm{M} \mathrm{HCl}$ (B) $0.01 \mathrm{M} \mathrm{NaOH}$ (C) $0.01 \mathrm{M} \mathrm{CH}_{3} \mathrm{COONa}$ (D) $0.01 \mathrm{M} \mathrm{NaCl}$(A) (C) (D) (B)(A) (D) (C) (B)$(\mathrm{B})(\mathrm{C})(\mathrm{D})(\mathrm{A})$(B) $$ (D) $$ (C) $$ (A)Correct Option: , 2 Solution: (A) $0.01 \mathrm{MHCl}$ $\left[\mathrm{H}^{+}\right]=10^{-2}, \mathrm{pH}=-\log 10^{-2}=2$ $\mathrm{pOH}=14-2=12$ (B) $0.01 \mathrm{...
Read More →Find the sum of all odd numbers between 0 and 50.
Question: Find the sum of all odd numbers between 0 and 50. Solution: All odd numbers between 0 and 50 are 1, 3, 5, 7, ..., 49. This is an AP in whicha= 1,d= (3 - 1) = 2 andl= 49. Let the number of terms ben.Then, Tn= 49⇒a+ (n- 1)d= 49⇒ 1 + (n- 1) ⨯ 2 = 49⇒ 2n=50⇒n=25 $\therefore$ Required sum $=\frac{n}{2}(a+l)$ $=\frac{25}{2}[1+49]=25 \times 25=625$ Hence, the required sum is 625....
Read More →If the sum and product of the first three terms in an A.P.
Question: If the sum and product of the first three terms in an A.P. are 33 and 1155 , respectively, then a value of its $11^{\text {th }}$ term is:(1) $-35$(2) 25(3) $-36$(4) $-25$Correct Option: , 4 Solution: Let three terms of A.P. are $a-d, a, a+d$ Sum of terms is, $a-d+a+a+d=33 \Rightarrow a=11$ Product bf terms is, $(a-d) a(a+d)=11\left(121-d^{2}\right)=1155$ $\Rightarrow 121-d^{2}=105 \Rightarrow d=\pm 4$ if $d=4$ $\mathrm{T}_{11}=\mathrm{T}_{1}+10 d=7+10(4)=47$ if $d=-4$ $\mathrm{T}_{11}...
Read More →Water flows in a horizontal tube (see figure).
Question: Water flows in a horizontal tube (see figure). The pressure of water changes by $700 \mathrm{Nm}^{-2}$ between $A$ and $B$ where the area of cross section are $40 \mathrm{~cm}^{2}$ and $20 \mathrm{~cm}^{2}$, respectively. Find the rate of flow of water through the tube. (density of water $=1000 \mathrm{kgm}^{-3}$ ) (1) $3020 \mathrm{~cm}^{3} / \mathrm{s}$(2) $2720 \mathrm{~cm}^{3} / \mathrm{s}$(3) $2420 \mathrm{~cm}^{3} / \mathrm{s}$(4) $1810 \mathrm{~cm}^{3} / \mathrm{s}$Correct Optio...
Read More →Solve the following
Question: For a reaction $\mathrm{X}+\mathrm{Y} \rightleftharpoons 2 \mathrm{Z}, 1.0 \mathrm{~mol}$ of $\mathrm{X}, 1.5 \mathrm{~mol}$ of $\mathrm{Y}$ and $0.5 \mathrm{~mol}$ of $Z$ were taken in a $1 \mathrm{~L}$ vessel and allowed to react. At equilibrium, the concentration of $\mathrm{Z}$ was $1.0 \mathrm{~mol} \mathrm{~L}^{-1}$. The equilibrium constant of the reaction is $\frac{x}{15}$. The value of $x$ is____________ . Solution: (16)...
Read More →How many terms of the AP 20,
Question: How many terms of the AP $20,19 \frac{1}{3}, 18 \frac{2}{3}, \ldots$ must be taken so that their sum is $300 ?$ Explain the double answer. Solution: The given AP is $20,19 \frac{1}{3}, 18 \frac{2}{3}, \ldots .$ Here, $a=20$ and $d=19 \frac{1}{3}-20=\frac{58}{3}-20=\frac{58-60}{3}=-\frac{2}{3}$ Let the required number of terms ben. Then, $S_{n}=300$ $\Rightarrow \frac{n}{2}\left[2 \times 20+(n-1) \times\left(-\frac{2}{3}\right)\right]=300 \quad\left\{S_{n}=\frac{n}{2}[2 a+(n-1) d]\right...
Read More →Let the sum of the first
Question: Let the sum of the first $\mathrm{n}$ terms of a non-constant A.P., $\mathrm{a}_{1}$, $a_{2}, a_{3}$, be $50 n+\frac{n(n-7)}{2} A$, where $A$ is a constant. If $\mathrm{d}$ is the common difference of this A.P., then the ordered pair $\left(\mathrm{d}, \mathrm{a}_{50}\right)$ is equal to:(1) $(50,50+46 \mathrm{~A})$(2) $(50,50+45 \mathrm{~A})$(3) $(\mathrm{A}, 50+45 \mathrm{~A})$(4) $(\mathrm{A}, 50+46 \mathrm{~A})$Correct Option: , 4 Solution: $\because S_{n}=\left(50-\frac{7 A}{2}\ri...
Read More →Two liquids of densities
Question: Two liquids of densities $\rho_{1}$ and $\rho_{2}\left(\rho_{2}=2 \rho_{1}\right)$ are filled up behind a square wall of side $10 \mathrm{~m}$ as shown in figure. Each liquid has a height of $5 \mathrm{~m}$. The ratio of the forces due to these liquids exerted on upper part $M N$ to that at the lower part $N O$ is (Assume that the liquids are not mixing):(1) $1 / 3$(2) $2 / 3$(3) $1 / 2$(4) $1 / 4$Correct Option: , 4 Solution: Let $P_{1}, P_{2}$ and $P_{3}$ be the pressure at points $M...
Read More →A soft drink was bottled with a partial prssure of
Question: A soft drink was bottled with a partial prssure of $\mathrm{CO}_{2}$ of 3 bar over the liquid at room temperature. The partial pressure of $\mathrm{CO}_{2}$ over the solution approaches a value of 30 bar when $44 \mathrm{~g}$ of $\mathrm{CO}_{2}$ is dissolved in $1 \mathrm{~kg}$ of water at room temperature. The approximate $\mathrm{pH}$ of the soft drink is_____________ $\times 10^{-1}$. (First dissociation constant of $\mathrm{H}_{2} \mathrm{CO}_{3}=4.0 \times 10^{-7} ; \log$ $2=0.3 ...
Read More →Prove the following
Question: The sum $\sum_{k=1}^{20} k \frac{1}{2^{k}}$ is equal to :(1) $2-\frac{3}{2^{17}}$(2) $1-\frac{11}{2^{20}}$(3) $2-\frac{11}{2^{19}}$(4) $2-\frac{21}{2^{20}}$Correct Option: , 3 Solution: Let, $S=\sum_{k=1}^{20} k \cdot \frac{1}{2^{k}}$ $S=\frac{1}{2}+2 \cdot \frac{1}{2^{2}}+3 \cdot \frac{1}{2^{3}}+\ldots .+20 \cdot \frac{1}{2^{20}}$......(i) $\frac{1}{2} S=\frac{1}{2^{2}}+2 \cdot \frac{1}{2^{3}}+\ldots+19 \frac{1}{2^{20}}+20 \frac{1}{2^{21}}$...(ii) On subtracting equations (ii) by (i),...
Read More →How many terms of the AP 63, 60, 57, 54, ...
Question: How many terms of the AP 63, 60, 57, 54, ... must be taken so that their sum is 693? Explain the double answer. Solution: The given AP is 63, 60, 57, 54, ... .Here,a= 63 andd= 60 63 = 3Let the required number of terms ben. Then, $S_{n}=693$ $\Rightarrow \frac{n}{2}[2 \times 63+(n-1) \times(-3)]=693 \quad\left\{S_{n}=\frac{n}{2}[2 a+(n-1) d]\right\}$ $\Rightarrow \frac{n}{2}(126-3 n+3)=693$ $\Rightarrow n(129-3 n)=1386$ $\Rightarrow 3 n^{2}-129 n+1386=0$ $\Rightarrow 3 n^{2}-66 n-63 n+1...
Read More →100 mL of 0.1MHCl is taken in a beaker and to it 100mL
Question: $100 \mathrm{~mL}$ of $0.1 \mathrm{MHCl}$ is taken in a beaker and to it $100 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{NaOH}$ is added in steps of $2 \mathrm{~mL}$ and the $\mathrm{pH}$ is continuously measured. Which of the following graphs correctly depicts the change in $\mathrm{pH}$ ?Correct Option: , 3 Solution: At equivalence point $\mathrm{pH}$ is 7 and $\mathrm{pH}$ increases with addition of $\mathrm{NaOH}$ so correct graph is (c)....
Read More →If three distinct numbers a, b, c are
Question: If three distinct numbers $a, b, c$ are in G.P. and the equations $a x^{2}+2 b x+c=0$ and $d x^{2}+2 e x+f=0$ have a common root, then which one of the following statements is correct?(1) $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in A.P.(2) $d, e, f$ are in A.P.(3) $d, e, f$ are in G.P.(4) $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in G.P.Correct Option: 1 Solution: Since $a, b, c$ are in G.P. $\therefore b^{2}=a c$ Given equation is, $a x^{2}+2 b x+c=0$ $\Rightarrow a x^{2}+2 \sqrt...
Read More →A leak proof cylinder of length 1 m,
Question: A leak proof cylinder of length $1 \mathrm{~m}$, made of a metal which has very low coefficient of expansion is floating vertically in water at $0^{\circ} \mathrm{C}$ such that its height above the water surface is $20 \mathrm{~cm}$. When the temperature of water is increased to $4^{\circ} \mathrm{C}$, the height of the cylinder above the water surface becomes $21 \mathrm{~cm}$. The density of water at $T=4^{\circ} \mathrm{C}$, relative to the density at $T=0^{\circ} \mathrm{C}$ is clo...
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