The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below:
Question: The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below: Concentration/M 0.001 0.010 0.020 0.050 0.100 102 /S m11.237 11.85 23.15 55.53 106.74 Calculate $\Lambda_{m}$ for all concentrations and draw a plot between $\Lambda_{m}$ and $\mathrm{c}^{1 / 2}$. Find the value of $\Lambda_{m}^{0}$. Solution: Given, = 1.237 102S m1, c = 0.001 M Then,= 1.237 104S cm1,c = 0.0316 M1/2 $\therefore \Lambda_{m}=\frac{\kappa}{c}$ $=\f...
Read More →The radius of a circle with centre O is 5 cm (Fig . 15.84).
Question: The radius of a circle with centreOis 5 cm (Fig . 15.84). Two radiiOAandOBare drawn at right angles to each other. Find the areas of the segments made by the chordAB (Take = 3.14). Solution: We have given that the radius of the circle is 5 cm. First we will find the area of the minor segment AB as given below, Area of the minor segment $\mathrm{AB}=\left(\frac{\pi \theta}{360}-\sin \frac{\theta}{2} \cos \frac{\theta}{2}\right) r^{2}$ Substituting the values we get, Area of the minor se...
Read More →The area of a sector is one-twelfth that of the complete circle.
Question: The area of a sector is one-twelfth that of the complete circle. Find the angle of the sector. Solution: We have given that area of the sector is one-twelfth of the area of the circle. $\therefore \frac{\theta}{360} \times \pi r^{2}=\frac{1}{12} \times \pi r^{2}$ Now we will simplify the above equation as below, $\therefore \frac{\theta}{360} \times \pi r^{2}=\frac{1}{12} \times \pi r^{2}$ $\therefore \frac{\theta}{360}=\frac{1}{12}$ Now we will multiply both sides of the equation by 3...
Read More →The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Ω.
Question: The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 103S cm1. Solution: Given, Conductivity,= 0.146 103S cm1 Resistance,R= 1500 Ω Cell constant =R = 0.146 103 1500 = 0.219 cm1...
Read More →From a circular piece of cardboard of radius 3 cm two sectors of 90° have been cut off.
Question: From a circular piece of cardboard of radius 3 cm two sectors of 90 have been cut off. Find the perimeter of the remaining portion nearest hundredth centimeters (Take = 22/7). Solution: It is very clear that if we cut two sectors of $90^{\circ}$ then we will be left with a semi-circle without diameter. Look at the figure. If we cut sectors OPA and OPB then we will get curve AB. Therefore, the required perimeter will be length of the curve AB. Now we will calculate the perimeter of the ...
Read More →The conductivity of 0.20 M solution of KCl at 298 K is
Question: The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 Scm1. Calculate its molar conductivity. Solution: Given, = 0.0248 S cm1 c = 0.20 M $\therefore$ Molar conductivity, $\Lambda_{m}=\frac{\kappa \times 1000}{\mathrm{c}}$ $=\frac{0.0248 \times 1000}{0.2}$ = 124 Scm2mol1...
Read More →In what ratio does the x-axis divide the join of A(2, −3) and B(5, 6)?
Question: In what ratio does thex-axis divide the join ofA(2, 3) andB(5, 6)?(a) 2 : 3(b) 3 : 5(c) 1 : 2(d) 2 : 1 Solution: (c) 1 : 2 Let $A B$ be divided by the $x$ axis in the ratio $k: 1$ at the point $P$. Then, by section formula, the coordinates ofPare $P\left(\frac{5 k+2}{k+1}, \frac{6 k-3}{k+1}\right)$ ButPlies on thexaxis: so, its ordinate is 0. $\frac{6 k-3}{k+1}=0$ $\Rightarrow 6 k-3=0$ $\Rightarrow 6 k=3$ $\Rightarrow k=\frac{1}{2}$ Hence, the required ratio is $\frac{1}{2}: 1$, which ...
Read More →The diameter of a coin is 1 cm (in the following figure).
Question: The diameter of a coin is 1 cm (in the following figure). If four such coins be placed on a table so that the rim of each touches that of the other two, find the area of the shaded region (Take = 3.1416). Solution: Look at the figure carefully shaded region is bounded between four sectors of the circle with same radius and a square of side 1 cm. Therefore, the area of the shaded region is nothing but the difference the area of the square and area of one circle. $\therefore$ Area of the...
Read More →Define conductivity and molar conductivity for the solution of an electrolyte .
Question: Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration. Solution: Conductivity of a solution is defined as the conductance of a solution of 1 cm in length and area of cross-section 1 sq. cm. The inverse of resistivity is called conductivity or specific conductance. It is represented by the symbol. If is resistivity, then we can write: $\kappa=\frac{1}{\rho}$ The conductivity of a solution at any given concentration is t...
Read More →The distance of the point P(3, 4) from the x-axis is
Question: The distance of the pointP(3, 4) from thex-axis is(a) 3 units(b) 4 units(c) 7 units(d) 1 unit Solution: (b) 4 unitsThey-coordinate is the distance of the point from thex-axis.Here, they-coordinate is 4....
Read More →In the following figure, ABCD is a rectangle,
Question: In the following figure,ABCDis a rectangle, havingAB= 20 cm andBC= 14 cm. Two sectors of 180 have been cut off. Calculate:(i) the area of the shaded region.(ii) the length of the boundary of the shaded region. Solution: (i) We have given two semi-circles and a rectangle. Area of the shaded region = Area of the rectangle Area of the two semicircles ..(1) $\therefore$ Area of the shaded region $=20 \times 14-2 \times \frac{1}{2} \times \pi \times 7 \times 7$ Substituting $\pi=\frac{22}{7...
Read More →In the button cells widely used in watches and other devices the following reaction takes place:
Question: In the button cells widely used in watches and other devices the following reaction takes place: Zn(s) + Ag2O(s) + H2O(l) Zn2+(aq) + 2Ag(s) + 2OH(aq) Determine $\Delta_{r} G^{\oplus}$ and $E^{\oplus}$ for the reaction. Solution: $\therefore E^{\ominus}=1.104 \mathrm{~V}$ We know that, $\Delta_{r} G^{\ominus}=-n \mathrm{~F} E^{\ominus}$ = 2 96487 1.04 = 213043.296 J = 213.04 kJ...
Read More →Which point on x-axis is equidistant from the points A(7, 6) and B(−3, 4)?
Question: Which point onx-axis is equidistant from the pointsA(7, 6) andB(3, 4)? (a) $(0,4)$ (b) $(-4,0)$ (c) $(3,0)$ (d) $(0,3)$ Solution: LetP(x, 0) be the point onx-axis. Then as per the question $A P=B P \Rightarrow A P^{2}=B P^{2}$ $\Rightarrow(x-7)^{2}+(0-6)^{2}=(x+3)^{2}+(0-4)^{2}$ $\Rightarrow x^{2}-14 x+49+36=x^{2}+6 x+9+16$ $\Rightarrow 60=20 x$ $\Rightarrow x=\frac{60}{20}=3$ Thus, the required point is (3, 0).Hence, the correct answer is option (c)....
Read More →Write the Nernst equation and emf of the following cells at 298 K:
Question: Write the Nernst equation and emf of the following cells at 298 K: (i)Mg(s) | Mg2+(0.001M) || Cu2+(0.0001 M) | Cu(s) (ii)Fe(s) | Fe2+(0.001M) || H+(1M)|H2(g)(1bar) | Pt(s) (iii)Sn(s) | Sn2+(0.050 M) || H+(0.020 M) | H2(g) (1 bar) | Pt(s) (iv)Pt(s) | Br2(l) | Br(0.010 M) || H+(0.030 M) | H2(g) (1 bar) | Pt(s). Solution: (i)For the given reaction, the Nernst equation can be given as: $E_{\text {cell }}=E_{\text {cell }}^{\ominus}-\frac{0.0591}{n} \log \frac{\left[\mathrm{Mg}^{2+}\right]}...
Read More →In the following figure, the square ABCD is divided into five equal parts,
Question: In the following figure, the squareABCDis divided into five equal parts, all having same area. The central part is circular and the linesAE,GC,BFandHDlie along the diagonalsACandBDof the square. IfAB= 22 cm, find:(i) the circumference of the central part.(ii) the perimeter of the partABEF. Solution: We have a square ABCD. We have, $\mathrm{AB}=22 \mathrm{~cm}$ (i)We have to find the perimeter of the triangle. We have a relation as, Area of circular region $=\frac{1}{5}($ Area of $\math...
Read More →If A(−6, 7) and B(−1, −5) are two given points then the distance 2AB is
Question: IfA(6, 7) andB(1, 5) are two given points then the distance 2ABis (a) 13 (b) 26 (c) 169 (d) 238 Solution: The given points areA(6, 7) andB(1, 5). So $A B=\sqrt{(-6+1)^{2}+(7+5)^{2}}$ $=\sqrt{(-5)^{2}+(12)^{2}}$ $=\sqrt{25+144}$ $=\sqrt{169}$ $=13$ Thus, 2AB= 26.Hence, the correct answer is option (b)....
Read More →Calculate the standard cell potentials of galvanic cells in which the following reactions take place:
Question: Calculate the standard cell potentials of galvanic cells in which the following reactions take place: (i)2Cr(s) + 3Cd2+(aq) 2Cr3+(aq) + 3Cd (ii)Fe2+(aq) + Ag+(aq) Fe3+(aq) + Ag(s) Calculate the ΔrGand equilibrium constant of the reactions. Solution: (i) $E_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{\ominus}=0.74 \mathrm{~V}$ $E_{\mathrm{Cd}^{2+} / \mathrm{Cd}}^{\ominus}=-0.40 \mathrm{~V}$ The galvanic cell of the given reaction is depicted as: $\mathrm{Cr}_{(s)}\left|\mathrm{Cr}^{3+}{ }_{(a q)}...
Read More →The point P which divides the line segment joining the points A(2, −5) and B(5, 2) in the ratio 2 : 3
Question: The pointPwhich divides the line segment joining the pointsA(2, 5) andB(5, 2) in the ratio 2 : 3 lies in the quadrant (a) I (b) II (c) III (d) IV Solution: Let (x,y) be the coordinates ofP. Then $x=\frac{2 \times 5+3 \times 2}{2+3}=\frac{10+6}{5}=\frac{16}{5}$ $y=\frac{2 \times 2+3 \times(-5)}{2+3}=\frac{4-15}{5}=\frac{-11}{5}$ Thus, the coordinates of point $P$ are $\left(\frac{16}{5}, \frac{-11}{5}\right)$ and so it lies in the fourth quadrant. Hence, the correct answer is option (d)...
Read More →A circular field has a perimeter of 650 m.
Question: A circular field has a perimeter of 650 m. A square plot having its vertices on the circumference of the field is marked in the field. Calculate the area of the square plot. Solution: We have a circular field in which a square field is marked. Let the radius of the circle ber. We have, Perimeter $=650$ $r=\frac{325}{\pi}$ Use Pythagoras theorem to find the side of square as, $\mathrm{AB}=\sqrt{r^{2}+r^{2}}$ $=\frac{325}{\pi} \sqrt{2}$ So area of the square plot, $=(\mathrm{AB})^{2}$ $=...
Read More →Calculate the standard cell potentials of galvanic cells in which the following reactions take place:
Question: Calculate the standard cell potentials of galvanic cells in which the following reactions take place: (i)2Cr(s) + 3Cd2+(aq) 2Cr3+(aq) + 3Cd (ii)Fe2+(aq) + Ag+(aq) Fe3+(aq) + Ag(s) Calculate the ΔrGand equilibrium constant of the reactions. Solution: The galvanic cell of the given reaction is depicted as: $\mathrm{Cr}_{(s)}\left|\mathrm{Cr}^{3+}{ }_{(a q)}\right|\left|\mathrm{Cd}^{2+}{ }_{(a q)}\right| \mathrm{Cd}_{(s)}$ Now, the standard cell potential is $E_{\text {cell }}^{\ominus}=E...
Read More →In the following figure, an equilateral triangle ABC
Question: In the following figure, an equilateral triangle ABC of side 6 cm has been inscribed in a circle. Find the area of the shaded region. (Take = 3.14). Solution: We have to find the area of the shaded portion. We have $\triangle \mathrm{ABC}$ which is an equilateral triangle and $\mathrm{AB}=6 \mathrm{~cm}$. Let $r$ be the radius of the circle. We have O as the circumcentre. $\angle \mathrm{OBA}=30^{\circ}$ So, $\cos \left(30^{\circ}\right)=\frac{3}{r}$ Thus, $r=2 \sqrt{3}$ So area of the...
Read More →The midpoint of segment AB is P(0, 4).
Question: The midpoint of segmentABisP(0, 4). If the coordinates ofBare (2, 3), then the coordinates ofAare (a) $(2,5)$ (b) $(-2,-5)$ (c) $(2,9)$ (d) $(-2,11)$ Solution: Let (x,y) be the coordinates ofA. Then $0=\frac{-2+x}{2} \Rightarrow x=2$ $4=\frac{3+y}{2} \Rightarrow y=8-3=5$ Thus, the coordinates ofAare (2, 5).Hence, the correct answer is option (a)....
Read More →In the given figure P(5, −3) and Q(3, y) are the points of trisection of the line segment
Question: In the given figureP(5, 3) andQ(3,y) are the points of trisection of the line segmentjoiningA(7, 2) andB(1, 5). Then,yequals (a) 2 (b) 4 (c) $-4$ (d) $\frac{-5}{2}$ Solution: Here,AQ:BQ= 2 : 1. Then $y=\frac{2 \times(-5)+1 \times(-2)}{2+1}$ $=\frac{-10-2}{3}$ $=-4$ Hence, the correct answer is option (c)....
Read More →A circle is inscribed in an equilateral triangle ABC is side 12 cm,
Question: A circle is inscribed in an equilateral triangle ABC is side 12 cm, touching its sides (the following figure). Find the radius of the inscribed circle and the area of the shaded part. Solution: We have to find the area of the shaded portion. We have $\triangle \mathrm{ABC}$ which is an equilateral triangle and $\mathrm{AB}=12 \mathrm{~cm}$. We have O as the incentre and OP, OQ and OR are equal. So, $a r(\Delta \mathrm{ABC})=a r(\Delta \mathrm{OAB})+a r(\Delta \mathrm{OBC})+a r(\Delta \...
Read More →Depict the galvanic cell in which the reaction
Question: Depict the galvanic cell in which the reaction Zn(s) + 2Ag+(aq) Zn2+(aq) + 2Ag(s) takes place. Further show: (i)Which of the electrode is negatively charged? (ii)The carriers of the current in the cell. (iii)Individual reaction at each electrode. Solution: The galvanic cell in which the given reaction takes place is depicted as: $\mathrm{Zn}_{(s)}\left|\mathrm{Zn}^{2+}{ }_{(a q)} \| \mathrm{Ag}_{(a q)}^{+}\right| \mathrm{Ag}_{(s)}$ (i)Zn electrode (anode) is negatively charged. (ii)Ion...
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