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?
eSaral Foundation courses How many terms of the AP 21, 18, 15, ... must be added to get the sum 0? 🚀 Checkout eSaral Courses What Is This AP and Why Does Its Sum Reach Zero? The sequence 21, 18, 15, is a decreasing arithmetic progression (AP). Each term is 3 less than the previous one, so the common difference d = 3. The terms start positive but steadily cross zero (the 8th term equals zero) and then become negative. Because positive early terms and negative later terms exist in the same sequenc...
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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