An acidic buffer is obtained on mixing :
Question: An acidic buffer is obtained on mixing :$100 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}$ and $100 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{NaOH}$$100 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{HCl}$ and $200 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{NaCl}$$100 \mathrm{~mL}$ of $0.1 \mathrm{MCH}_{3} \mathrm{COOH}$ and $200 \mathrm{~mL}$ of $0.1 \mathrm{MNaOH}$$100 \mathrm{~mL}$ of $0.1 \mathrm{MHCl}$ and $200 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COON...
Read More →The number of terms common to the two
Question: The number of terms common to the two A.P.'s $3,7,11, \ldots$, 407 and $2,9,16, \ldots, 709$ is__________. Solution: First common term of both the series is 23 and common difference is $7 \times 4=28$ $\because \quad$ Last term $\leq 407$ $\Rightarrow \quad 23+(n-1) \times 28 \leq 407$ $\Rightarrow \quad(n-1) \times 28 \leq 384$ $\Rightarrow \quad n \leq \frac{384}{28}+1$ $\Rightarrow n \leq 14.71$ Hence, $n=14$...
Read More →How many terms of the AP 9, 17, 25, ... must be taken so that their sum is 636?
Question: How many terms of the AP 9, 17, 25, ... must be taken so that their sum is 636? Solution: The given AP is 9, 17, 25, ... .Here,a= 9 andd= 17 9 = 8Let the required number of terms ben. Then, $S_{n}=636$ $\Rightarrow \frac{n}{2}[2 \times 9+(n-1) \times 8]=636 \quad\left\{S_{n}=\frac{n}{2}[2 a+(n-1) d]\right\}$ $\Rightarrow \frac{n}{2}(18+8 n-8)=636$ $\Rightarrow \frac{n}{2}(10+8 n)=636$ $\Rightarrow n(5+4 n)=636$ $\Rightarrow 4 n^{2}+5 n-636=0$ $\Rightarrow 4 n^{2}-48 n+53 n-636=0$ $\Rig...
Read More →Prove the following
Question: Let $a_{n}$ be the $n^{\text {th }}$ term of a G.P. of positive terms. If $\sum_{n=1}^{100} a_{2 n+1}=200$ and $\sum_{n=1}^{100} a_{2 n}=100$, then $\sum_{n=1}^{200} a_{n}$ is equal to :(1) 300(2) 225(3) 175(4) 150Correct Option: , 4 Solution: Let G.P. be $a, a r, a r^{2}$ $\sum_{n=1}^{100} a_{2 n+1}=a_{3}+a_{5}+\ldots . .+a_{201}=200$ $\Rightarrow \frac{a r^{2}\left(r^{200}-1\right)}{r^{2}-1}=200$ ...(i) $\sum_{n=1}^{100} a_{2 n}=a_{2}+a_{4}+\ldots . .+a_{200}=100$ $\Rightarrow \quad ...
Read More →For the following Assertion and Reason, the correct option is
Question: For the following Assertion and Reason, the correct option is Assertion (A): When $\mathrm{Cu}$ (II) and sulphide ions are mixed, they react together extremely quickly to give a solid. Reason (R): The equilibrium constant of $\mathrm{Cu}^{2+}(\mathrm{aq})+$ $\mathrm{S}^{2-}(\mathrm{aq}) \rightleftharpoons \mathrm{CuS}(\mathrm{s})$ is high because the solubility product is low.(A) is false and (R) is trueBoth $(\mathbf{A})$ and $(\mathbf{R})$ are falseBoth (A) and (R) are true but (R) i...
Read More →The pH of ammonium phosphate solution,
Question: The $\mathrm{pH}$ of ammonium phosphate solution, if $\mathrm{pk}_{\mathrm{a}}$ of phosphoric acid and $\mathrm{pk}_{\mathrm{b}}$ of ammonium hydroxide are $5.23$ and $4.75$ respectively, is Solution: (7)...
Read More →The product
Question: The product $2^{\frac{1}{4}} \cdot 4^{\frac{1}{16}} \cdot 8^{\frac{1}{48}} \cdot 16^{\frac{1}{128}} \ldots$ to $\infty$ is equal to:(1) $2^{\frac{1}{2}}$(2) $2^{\frac{1}{4}}$(3) 1(4) 2Correct Option: 1 Solution: $2^{\frac{1}{4}}+\frac{2}{16}+\frac{3}{48}+\ldots \ldots \infty$ $=2^{\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots \ldots \infty}=\sqrt{2}$...
Read More →Prove the following
Question: The sum, $\sum_{n=1}^{7} \frac{n(n+1)(2 n+1)}{4}$ is equal to__________. Solution: $\left[\sum_{n=1}^{7} \frac{n(n+1)(2 n+1)}{4}\right] \frac{1}{4}\left[\sum_{n=1}^{7}\left(2 n^{3}+3 n^{2}+n\right)\right]$ $=\frac{1}{4}\left(2\left(\frac{7.8}{2}\right)^{2}+3\left(\frac{7.8 .15}{6}\right)+\frac{7.8}{2}\right)$ $\Rightarrow \quad \frac{1}{4}[2 \times 49 \times 16+28 \times 15+28]$ $=\frac{1}{4}[1568+420+28]=504$...
Read More →The solubility of
Question: The solubility of $\mathrm{Ca}(\mathrm{OH})_{2}$ in water is : [Given : The solubility product of $\mathrm{Ca}(\mathrm{OH})_{2}$ in water $=5.5 \times 10^{-6}$ ]$1.11 \times 10^{-6}$$1.77 \times 10^{-6}$$1.77 \times 10^{-2}$$1.11 \times 10^{-2}$Correct Option: , 4 Solution:...
Read More →How many terms of the AP 21, 18, 15, ... must be added to get the sum 0?
Question: How many terms of the AP 21, 18, 15, ... must be added to get the sum 0? Solution: The given AP is 21, 18, 15, ... .Here,a= 21 andd= 18 21 = 3Let the required number of terms ben. Then, $S_{n}=0$ $\Rightarrow \frac{n}{2}[2 \times 21+(n-1) \times(-3)]=0 \quad\left\{S_{n}=\frac{n}{2}[2 a+(n-1) d]\right\}$ $\Rightarrow \frac{n}{2}(42-3 n+3)=0$ $\Rightarrow n(45-3 n)=0$ $\Rightarrow n=0$ or $45-3 n=0$ $\Rightarrow n=0$ or $n=15$ n= 15 (Number of terms cannot be zero)Hence, the required num...
Read More →If mth term of an AP is
Question: If $m$ th term of an AP is $\frac{1}{n}$ and $n$th term is $\frac{1}{m}$ then find the sum of its first $m n$ terms. Solution: Supposeabe the first term andd be the common difference of the given AP. $a_{m}=\frac{1}{n}$ $\Rightarrow a+(m-1) d=\frac{1}{n} \quad \ldots \ldots(1)$ And, $a_{n}=\frac{1}{m}$ $\Rightarrow a+(n-1) d=\frac{1}{m} \quad \ldots .(2)$ Subtracting (2) from (1), we get $\frac{1}{n}-\frac{1}{m}=(m-n) d$ $\Rightarrow \frac{m-n}{m n}=(m-n) d$ $\Rightarrow d=\frac{1}{m n...
Read More →The solubility of AgCN in a buffer solution of pH}=3 is X.
Question: The solubility of $\mathrm{AgCN}$ in a buffer solution of $\mathrm{pH}=3$ is $\mathrm{x} .$ The value of $\mathrm{X}$ is: [Assume: No cyano complex is formed; $\mathrm{K}_{\mathrm{sp}}(\mathrm{AgCN})=2.2 \times 10^{-16}$ and $\mathrm{K}_{\mathrm{a}}(\mathrm{HCN})=6.2 \times 10^{-10}$ ]$0.625 \times 10^{-6}$$1.6 \times 10^{-6}$$2.2 \times 10^{-16}$$1.9 \times 10^{-5}$Correct Option: , 4 Solution:...
Read More →Prove the following
Question: If the $10^{\text {th }}$ term of an A.P. is $\frac{1}{20}$ and its $20^{\text {th }}$ term is $\frac{1}{10}$, then the sum of its first 200 terms is:(1) 50(2) $50 \frac{1}{4}$(3) 100(4) $100 \frac{1}{2}$Correct Option: , 4 Solution: $T_{10}=\frac{1}{20}=a+9 d$...(i) $T_{20}=\frac{1}{10}=a+19 d$...(ii) Solving equations (i) and (ii), we get $a=\frac{1}{200}, d=\frac{1}{200}$ $\Rightarrow \quad S_{200}=\frac{200}{2}\left[\frac{2}{200}+\frac{199}{200}\right]=\frac{201}{2}=100 \frac{1}{2}...
Read More →The sum
Question: The sum $\sum_{k=1}^{20}(1+2+3+\ldots+k)$ is________. Solution: Given series can be written as $\sum_{k=1}^{20} \frac{k(k+1)}{2}=\frac{1}{2} \sum_{k=1}^{20}\left(k^{2}+k\right)$ $=\frac{1}{2}\left[\frac{20(21)(41)}{6}+\frac{20(21)}{2}\right]$ $=\frac{1}{2}\left[\frac{420 \times 41}{6}+\frac{20 \times 21}{2}\right]=\frac{1}{2}[2870+210]=1540$...
Read More →The solubility product of
Question: The solubility product of $\mathrm{PbI}_{2}$ is $8.0 \times 10^{-9}$. The solubility of lead iodide in $0.1$ molar solution of lead nitrate is $\mathrm{x} \times 10^{-6} \mathrm{~mol} / \mathrm{L}$. The value of $\mathrm{x}$ is _______________.(Rounded off to the nearest integer) [ Given $\sqrt{2}=1.41$ ] Solution: (141) $\mathrm{K}_{\mathrm{SP}}\left(\mathrm{PbI}_{2}\right)=8 \times 10^{-9}$ $\mathrm{PbI}_{2}(\mathrm{~s}) \rightleftharpoons \mathrm{Pb}^{+2}(\mathrm{aq})+2 \mathrm{I}^{...
Read More →Prove the following
Question: Let $f: R \rightarrow R$ be such that for all $x \in R,\left(2^{1+x}+2^{1-x}\right), f(x)$ and $\left(3^{x}+3^{-x}\right)$ are in A.P., then the minimum value of $f(x)$ is:__________. Solution: If $2^{1-x}+2^{1+x}, f(x), 3^{x}+3^{-x}$ are in A.P., then $f(x)=\left(\frac{2^{1+x}+2^{1-x}+3^{x}+3^{-x}}{2}\right)$ $2 f(x)=2\left(2^{x}+\frac{1}{2^{x}}\right)+\left(3^{x}+\frac{1}{3^{x}}\right)$ Using AM $\geq$ GM $f(x) \geq 3$...
Read More →If the sum of the first 40 terms of the series,
Question: If the sum of the first 40 terms of the series, $3+4+8+9+$ $13+14+18+19+\ldots$ is $(102) \mathrm{m}$, then $\mathrm{m}$ is equal to:(1) 20(2) 25(3) 5(4) 10Correct Option: 1 Solution: $S=\underbrace{3+4}+\underbrace{8+9}+\underbrace{13+14}+\underbrace{18+19} \ldots . .40$ terms $S=7+17+27+37+47+\ldots .20$ terms $S_{40}=\frac{20}{2}[2 \times 7+(19) 10]=10[14+190]$ $=10[2040]=(102)(20)$ $\Rightarrow \quad m=20$...
Read More →Solve the following
Question: The solubility of $\mathrm{CdSO}_{4}$ in water is $8.0 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$. Its solubility in $0.01 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}$ solution is______________. (Round off to the Nearest integer) (Assume that solubility is much less than $0.01 \mathrm{M}$ ) Solution: (64) In pure water, $K_{s p}=S^{2}=\left(8 \times 10^{-4}\right)^{2}$ $=64 \times 10^{-8}$ In $0.01 \mathrm{MH}_{2} \mathrm{SO}_{4}$ $\mathrm{K}_{\mathrm{sp}}=\mathrm{x}(\mathrm{x}+0.01)...
Read More →Let a1, a2, a3,.......... be a G. P.
Question: Let $a_{1}, a_{2}, a_{3}, \ldots$ be $a$ G. P. such that $a_{1}0, a_{1}+a_{2}=4$ and $a_{3}+a_{4}=16$. If $\sum_{i=1}^{9} a_{i}=4 \lambda$, then $\lambda$ is equal to:(1) $-513$(2) $-171$(3) 171(4) $\frac{511}{3}$Correct Option: , 2 Solution: Since, $a_{1}+a_{2}=4 \Rightarrow a_{1}+a_{1} r=4$ ...(i) $a_{3}+a_{4}=16 \Rightarrow a_{1} r^{2}+a_{1} r^{3}=16$...(ii) From eqn. (i), $a_{1}=\frac{4}{1+r}$ and substituting the value of $a_{1}$, in eqn (ii), $\left(\frac{4}{1+r}\right)^{r^{2}}+\...
Read More →In order to prepare a buffer solution of
Question: In order to prepare a buffer solution of $\mathrm{pH} 5.74$ sodium acetate is added to acetic acid. If the concentration of acetic acid in the buffer is $1.0$ M, the concentration of sodium acetate in the buffer is_________________ M.(Round off to the Nearest Integer). [Given:pKa (acetic acid) $=4.74]$ Solution: (10) $\mathrm{pH}=\mathrm{pKa}+\log \frac{[\mathrm{CB}]}{[\mathrm{WA}]}$ $5.74=4.74+\log \frac{[\mathrm{CB}]}{1}$ $\Rightarrow[\mathrm{CB}]=10 \mathrm{M}$...
Read More →The sum of first n terms of an AP is
Question: (i) The sum of first $n$ terms of an AP is $\left(\frac{5 n^{2}}{2}+\frac{3 n}{2}\right)$. Find the $n$th term and the 20 th term of this AP. (ii) The sum of the first $n$ terms of an $\mathrm{AP}$ is $\left(\frac{3 n^{2}}{2}+\frac{5 n}{2}\right)$. Find its $n$th term and the 25 th term. Solution: (i) $s_{n}=\frac{5 n^{2}}{2}+\frac{3 n}{2}$ Sum of 1 term $=5\left(\frac{1}{2}\right)+3\left(\frac{1}{2}\right)=4$ Sum of 2 term $s=5\left(\frac{4}{2}\right)+3\left(\frac{2}{2}\right)=13$ 2 n...
Read More →Solve the following
Question: $0.01$ moles of a weak acid $\mathrm{HA}\left(\mathrm{K}_{\mathrm{a}}=2.0 \times 10^{-6}\right)$ is dissolved in $1.0 \mathrm{~L}$ of $0.1 \mathrm{MHCl}$ solution. The degree of dissociation of HA is______________. $\times 10^{-5}$ (Round off to the Nearest Integer). [Neglect volume change on adding HA. Assume degree of dissociation $1]$ Solution: (2) $\mathrm{HA} \rightleftharpoons \mathrm{H}^{+}+\mathrm{A}^{-}$ Initial conc. $0.01 \mathrm{M} \quad 0.1 \mathrm{M} \quad 0$ Equ. conc. $...
Read More →Five numbers are in A.P., whose sum is 25
Question: Five numbers are in A.P., whose sum is 25 and product is 2520 . If one of these five numbers is $-\frac{1}{2}$, then the greatest number amongst them is:(1) 27(2) 7(3) $\frac{21}{2}$(4) 16Correct Option: , 4 Solution: Let 5 terms of A.P. be $a-2 d, a-d, a, a+d, a+2 d$ Sum $=25 \Rightarrow 5 a=25 \Rightarrow a=5$ Product $=2520$ $(5-2 d)(5-d) 5(5+d)(5+2 d)=2520$ $\Rightarrow \quad\left(25-4 d^{2}\right)\left(25-d^{2}\right)=504$ $\Rightarrow \quad 625-100 d^{2}-25 d^{2}+4 d^{4}=504$ $\R...
Read More →Solve the following
Question: Sulphurous acid $\left(\mathrm{H}_{2} \mathrm{SO}_{3}\right)$ has $\mathrm{Ka}_{1}=1.7 \times 10^{-2}$ and $\mathrm{Ka}_{2}=6.4 \times 10^{-8} .$ The $\mathrm{pH}$ of $0.588 \mathrm{MH}_{2} \mathrm{SO}_{3}$ is_______________. (Round off to the Nearest Integer) Solution: (1) $\mathrm{H}_{2} \mathrm{SO}_{3}[$ Dibasic acid $] \mathrm{c}=0.588 \mathrm{M}$ $\Rightarrow \mathrm{pH}$ of solution $\mathrm{p}$ due to First dissociation only since $\mathrm{K}_{\mathrm{a}},\mathrm{Ka}_{2}$ $\Righ...
Read More →Suppose that a function
Question: Suppose that a function $f: \mathrm{R} \rightarrow \mathrm{R}$ satisfies $f(x+y)=f(x) f(y)$ for all $x, y \in \mathrm{R}$ and $f(1)=3$. If $\sum_{i=1}^{\mathrm{n}} f(i)=363$, then $\mathrm{n}$ is equal to_________. Solution: $\because f(x+y)=f(x) \cdot f(y) \quad \forall x \in \mathrm{R}$ and $f(1)=3$ $\Rightarrow f(x)=3^{x} \Rightarrow f(i)=3^{i}$ $\Rightarrow \sum_{i=1}^{n} f(i)=363 \Rightarrow 3+3^{2}+3^{3}+\ldots .+3^{n}=363$ $\Rightarrow \frac{3\left(3^{n}-1\right)}{3-1}=363 \quad...
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