The sum of the first 9 terms of an AP is 81 and that of its first 20 terms is 400.
Question: The sum of the first 9 terms of an AP is 81 and that of its first 20 terms is 400. Find the first term and the common difference of the AP. Solution: Sum of 9 term $s$ of $\mathrm{AP}=81$ Sum of $n$ terms of $\mathrm{AP}$ is $\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$ $81=\frac{9}{2}[2 a+8 d]$ $2 a+8 d=18$ $\ldots \ldots(1)$ Sum of 1 st 20 terms is 400 $400=\frac{20}{2}[2 a+19 d]$ $2 a+19 d=40$ $\ldots \ldots(2)$ Subtract (1) from (2) $d=2$ and $a=1$ Therefore, first term is 1 and common...
Read More →The oxygen dissolved in water exerts a partial pressure of 20 kPa in the vapour above water.
Question: The oxygen dissolved in water exerts a partial pressure of $20 \mathrm{kPa}$ in the vapour above water. The molar solubility of oxygen in water is __________ $\times 10^{-5} \mathrm{moldm}^{-3}$ (Round off to the Nearest Integer). [Given : Henry's law constant $=\mathrm{K}_{\mathrm{H}}=8.0 \times 10^{4} \mathrm{kPa}$ for $\mathrm{O}_{2}$ Density of water with dissolved oxygen $=1.0 \mathrm{~kg} \mathrm{dm}^{-3}$ ] Solution: (25) $\mathrm{P}_{-}\{(\mathrm{g})\}=\backslash$ left $\left[\...
Read More →The sum of the first 7 terms of an AP is 182.
Question: The sum of the first 7 terms of an AP is 182. If its 4th and 17th terms are in the ratio 1 : 5, find the AP. Solution: Letabe the first term anddbe the common difference of the AP. $\therefore S_{7}=182$ $\Rightarrow \frac{7}{2}(2 a+6 d)=182 \quad\left\{S_{n}=\frac{n}{2}[2 a+(n-1) d]\right\}$ $\Rightarrow a+3 d=26 \quad \ldots(1)$ Also, $a_{4}: a_{17}=1: 5$ (Given) $\Rightarrow \frac{a+3 d}{a+16 d}=\frac{1}{5} \quad\left[a_{n}=a+(n-1) d\right]$ $\Rightarrow 5 a+15 d=a+16 d$ $\Rightarro...
Read More →At 363K, the vapour pressure of A is 21kPa and that of B is 18 kPa.
Question: At $363 \mathrm{~K}$, the vapour pressure of $\mathrm{A}$ is $21 \mathrm{kPa}$ and that of $\mathrm{B}$ is $18 \mathrm{kPa}$. One mole of $\mathrm{A}$ and 2 moles of $\mathrm{B}$ are mixed. Assuming that this solution is ideal, the vapour pressure of the mixture is____________ $\mathrm{kPa}$. (Round of to the Nearest Integer). Solution: (19) Given $\mathrm{P}_{\mathrm{A}}^{0}=21 \mathrm{kPa} \quad \Rightarrow \mathrm{P}_{\mathrm{B}}^{0}=18 \mathrm{kPa}$ $\rightarrow$ An Ideal solution ...
Read More →Solve the following
Question: $\mathrm{AB}_{2}$ is $10 \%$ dissociated in water to $\mathrm{A}^{2+}$ and $\mathrm{B}^{-}$. The boiling point of a $10.0$ molal aqueous solution of $\mathrm{AB}_{2}$ is______________ . ${ }^{\circ} \mathrm{C}$. (Round off to the Nearest Integer). [Given : Molal elevation constant of water $\mathrm{K}_{\mathrm{b}}=0.5 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ boiling point of pure water $\left.=100^{\circ} \mathrm{C}\right]$ Solution: (106)...
Read More →The 12th term of an AP is −13 and the sum of its first four terms is 24.
Question: The 12th term of an AP is 13 and the sum of its first four terms is 24. Find the sum of its first 10 terms. Solution: Letabe the first term anddbe the common difference of the AP. Then, $a_{12}=-13$ $\Rightarrow a+11 d=-13 \quad \ldots \ldots(1) \quad\left[a_{n}=a+(n-1) d\right]$ Also, $S_{4}=24$ $\Rightarrow \frac{4}{2}(2 a+3 d)=24 \quad\left\{S_{n}=\frac{n}{2}[2 a+(n-1) d]\right\}$ $\Rightarrow 2 a+3 d=12 \quad \ldots .(2)$ Solving (1) and (2), we get $2(-13-11 d)+3 d=12$ $\Rightarro...
Read More →In an AP the first term is 22, nth term is −11 and sum to first nth terms is 66.
Question: In an AP the first term is 22,nth term is 11 and sum to firstnth terms is 66. Findnandd, the common difference. Solution: Here,a= 22, Tn=-11 and Sn= 66Letdbe the common difference of the given AP.Then Tn= -11⇒a+ (n- 1)d= 22 + (n- 1)d= -11⇒ (n- 1)d= -33 ...(i)The sum ofn terms of an AP is given by $S_{n}=\frac{n}{2}[2 a+(n-1) d]=66$ [Substituting the value of $(n-1) d$ from (i)] $\Rightarrow \frac{n}{2}[2 \times 22+(-33)]=\left(\frac{n}{2}\right) \times 11=66$ $\Rightarrow n=12$ Putting...
Read More →Which of the following compounds is likely to show
Question: Which of the following compounds is likely to show both Frenkel and Schottky defects in its crystalline form?$\mathrm{AgBr}$$\mathrm{CsCl}$$\mathrm{KBr}$$\mathrm{ZnS}$Correct Option: 1 Solution: AgBr shows both Schottky as well as Frenkel defects....
Read More →The first and the last terms of an AP are 5 and 45 respectively.
Question: The first and the last terms of an AP are 5 and 45 respectively. If the sum of all its terms is 400, find the common difference and the number of terms. Solution: Suppose there arenterms in the AP.Here,a= 5,l= 45 andSn= 400 $S_{n}=400$ $\Rightarrow \frac{n}{2}(5+45)=400 \quad\left[S_{n}=\frac{n}{2}(a+l)\right]$ $\Rightarrow \frac{n}{2} \times 50=400$ $\Rightarrow n=\frac{400 \times 2}{50}=16$ Thus, there are 16 terms in the AP.Letdbe the common difference of the AP. $\therefore a_{16}=...
Read More →A crystal is made up of metal ions '
Question: A crystal is made up of metal ions ' $\mathrm{M}_{1}$ ' and ' $\mathrm{M}_{2}$ ' and oxide ions. Oxide ions. form a $c c p$ lattice structure. The cation ' $\mathrm{M}_{1}$ ' occpies $50 \%$ of octahedral voids and the cation ' $\mathrm{M}_{2}$ ' occupies $12.5 \%$ of tetrahedral voids of oxide lattice. The oxidation numbers of ' $\mathrm{M}_{1}$ ' and ' $\mathrm{M}_{2}$ ' are, respectively:$+2,+4$$+1,+3$$+3,+1$$+4,+2$Correct Option: 1 Solution:...
Read More →The first and the last terms of an AP are 17 and 350 respectively.
Question: The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum? Solution: Suppose there arenterms in the AP.Here,a= 17,d= 9 andl= 350 $\therefore a_{n}=350$ $\Rightarrow 17+(n-1) \times 9=350 \quad\left[a_{n}=a+(n-1) d\right]$ $\Rightarrow 9 n+8=350$ $\Rightarrow 9 n=350-8=342$ $\Rightarrow n=38$ Thus, there are 38 terms in the AP. $\therefore S_{38}=\frac{38}{2}(17+350) \quad\left[S_{n}=\frac{n}{2}(a+l)\...
Read More →An element crystallises in a face-centred cubic (fcc) unit cell with cell edge a.
Question: An element crystallises in a face-centred cubic $(f c c)$ unit cell with cell edge $a$. The distance between the centres of two nearest octahedral voids in the crystal lattice is :$\frac{a}{\sqrt{2}}$a$\sqrt{2} a$$\frac{a}{2}$Correct Option: 1 Solution: Distance between two octahedral voids $x=\sqrt{\left(\frac{a}{2}\right)^{2}+\left(\frac{a}{2}\right)^{2}}=\sqrt{\frac{a^{2}}{4}+\frac{a^{2}}{4}}=\frac{a}{\sqrt{2}}$...
Read More →In an AP, the first term is −4, the last term is 29 and the sum of all its terms is 150.
Question: In an AP, the first term is 4, the last term is 29 and the sum of all its terms is 150. Find its common difference. Solution: Suppose there arenterms in the AP.Here,a= 4,l= 29 andSn= 150 $S_{n}=150$ $\Rightarrow \frac{n}{2}(-4+29)=150 \quad\left[S_{n}=\frac{n}{2}(a+l)\right]$ $\Rightarrow n=\frac{150 \times 2}{25}=12$ Thus, the AP contains 12 terms.Letdbe the common difference of the AP. $\therefore a_{12}=29$ $\Rightarrow-4+(12-1) \times d=29 \quad\left[a_{n}=a+(n-1) d\right]$ $\Right...
Read More →A diatomic molecule
Question: A diatomic molecule $\mathrm{X}_{2}$ has a body-centred cubic $(b c c)$ structure with a cell edge of $300 \mathrm{pm}$. The density of the molecule is $6.17 \mathrm{~g} \mathrm{~cm}^{-3}$. The number of molecules present in $200 \mathrm{~g}$ of $X_{2}$ is : (Avogadroconstant $\left.\left(\mathrm{N}_{\mathrm{A}}\right)=6 \times 10^{23} \mathrm{~mol}^{-1}\right)$$40 \mathrm{~N}_{\mathrm{A}}$$8 \mathrm{~N}_{\mathrm{A}}$$4 \mathrm{~N}_{\mathrm{A}}$$2 \mathrm{~N}_{\mathrm{A}}$Correct Optio...
Read More →Match the following:
Question: Match the following: (i)-(B), (ii)-(C), (iii)-(E), (iv)-(D)(i)-(D), (ii)-(B), (iii)-(A), (iv)-(E)(i)-(E), (ii)-(C), (iii)-(A), (iv)-(F)(i)-(D), (ii)-(B), (iii)-(E), (iv)-(F)Correct Option: , 3 Solution:...
Read More →In an AP the first term is 2, the last term is 29 and sum of all the terms is 155.
Question: In an AP the first term is 2, the last term is 29 and sum of all the terms is 155. Find the common difference of the AP. Solution: Here,a= 2,l=29 andSn= 155Letdbe the common difference of the given AP andnbe the total number of terms.Then Tn= 29⇒a+ (n- 1)d =29 ⇒ 2 + (n- 1)d =29 ...(i)The sum ofnterms of an AP is given by $S_{n}=\frac{n}{2}[a+l]=155$ $\Rightarrow \frac{n}{2}[2+29]=\left(\frac{n}{2}\right) \times 31=155$ $\Rightarrow n=10$ Putting the value ofnin (i), we get:⇒ 2 + 9d= 2...
Read More →An element with molar mass
Question: An element with molar mass $2.7 \times 10^{-2} \mathrm{~kg} \mathrm{~mol}^{-1}$ forms a cubic unit cell with edge length $405 \mathrm{pm}$. If its density is $2.7$ $\times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$, the radius of the element is approximately____________ . $\times 10^{-12} \mathrm{~m}$ (to the nearest integer). Solution: (143) $d=\frac{\mathrm{Z} \times \mathrm{M}}{\mathrm{N}_{\mathrm{A}} \times \text { Volume }}$ $2.7=\frac{\mathrm{Z} \times 27}{6.02 \times 10^{23} \times\l...
Read More →In an AP, it is given that
Question: In an $\mathrm{AP}$, it is given that $S_{5}+S_{7}=167$ and $S_{10}=235$, then find the $\mathrm{AP}$, where $S_{n}$ denotes the sum of its first $n$ terms. Solution: $S_{5}+S_{7}=167$ $\Rightarrow \frac{5}{2}(2 a+4 d)+\frac{7}{2}(2 a+6 d)=167 \quad\left\{S_{n}=\frac{n}{2}[2 a+(n-1) d]\right\}$ $\Rightarrow 5 a+10 d+7 a+21 d=167$ $\Rightarrow 12 a+31 d=167$ $\ldots \ldots(1)$ Also, $S_{10}=235$ $\Rightarrow \frac{10}{2}(2 a+9 d)=235$ $\Rightarrow 5(2 a+9 d)=235$ $\Rightarrow 2 a+9 d=47...
Read More →The number of octahedral voids per lattice site in a lattice is
Question: The number of octahedral voids per lattice site in a lattice is _________________ . (Rounded off to the nearest integer) Solution: (1) Assuming FCC No of lactice sites $=6$ face centre $+8$ corner $=14$ No. of octahedral voids $=13$ Ratio $=\frac{13}{14}=0.92857=1$ (Nearest integer)...
Read More →Find the sum of n terms of the following series:
Question: Find the sum ofnterms of the following series: $\left(4-\frac{1}{n}\right)+\left(4-\frac{2}{n}\right)+\left(4-\frac{3}{n}\right)+\ldots$ Solution: Let the given series be $X=\left(4-\frac{1}{n}\right)+\left(4-\frac{2}{n}\right)+\left(4-\frac{3}{n}\right)+\ldots$ $=[4+4+4+\ldots]-\left[\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+\ldots\right]$ $=4[1+1+1+\ldots]-\frac{1}{n}[1+2+3+\ldots]$ $=S_{1}-S_{2}$ $S_{1}=4[1+1+1+\ldots]$ $a=1, d=0$ $S_{1}=4 \times \frac{n}{2}[2 \times 1+(n-1) \times 0] \qu...
Read More →Find the sum of first 100 even natural numbers which are divisible by 5.
Question: Find the sum of first 100 even natural numbers which are divisible by 5. Solution: The first few even natural numbers which are divisible by 5are 10, 20, 30, 40, ... This is an AP in whicha= 10,d= (20 10) = 10 andn= 100 The sum ofn terms of an AP is given by $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ $=\left(\frac{100}{2}\right) \times[2 \times 10+(100-1) \times 10] \quad[\because a=10, d=10$ and $n=100]$ $=50 \times[20+990]=50 \times 1010=50500$ Hence, the sum of the first hundred even natural ...
Read More →The unit cell of copper corresponds to a face centered cube of edge length
Question: The unit cell of copper corresponds to a face centered cube of edge length 3.596 with one copper atom at each lattice point. The calculated density of copper in $\mathrm{kg} / \mathrm{m}^{3}$ is_______________ . [Molar mass of Cu: $63.54 \mathrm{~g} ;$ Avogadro number $\left.=6.022 \times 10^{23}\right]$ Solution: (9077) $\mathrm{d}=\frac{Z \times G M M}{N_{A} \times a^{3}}$ $\mathrm{d}=\frac{4 \times 63.54 \times 10^{-3}}{6.022 \times 10^{23} \times\left(3.596 \times 10^{-10}\right)^{...
Read More →The coordination number of an atom in a body-
Question: The coordination number of an atom in a body-centered cubic structure is _______________________ [Assume that the lattice is made up of atoms] Solution: (4) Fact...
Read More →In a binary compound, atoms of element A
Question: In a binary compound, atoms of element $\mathrm{A}$ form a hcp structure and those of element $\mathrm{M}$ occupy $2 / 3$ of the tetrahedral voids of the hep structure. The formula of the binary compound is:$\mathrm{M}_{2} \mathrm{~A}_{3}$$\mathrm{M}_{4} \mathrm{~A}_{3}$$\mathrm{M}_{4} \mathrm{~A}$$\mathrm{MA}_{3}$ Official Ans. by NTA (2)Correct Option: , 2 Solution: $\mathrm{M}_{12 \times \frac{2}{3}} \mathrm{~A}_{6}$ $\mathrm{M}_{8} \mathrm{~A}_{6}$ $\mathrm{M}_{4} \mathrm{~A}_{3}$...
Read More →A colloidal system consisting of a gas dispersed in
Question: A colloidal system consisting of a gas dispersed in a solid is called a/an:solid solgelaerosolfoamCorrect Option: 1 Solution: Colloid of gas dispersed in solid is called solid sol....
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