The equation of the normal to the curve
Question: The equation of the normal to the curve $y=x(2-x)$ at the point $(2,0)$ is A. $x-2 y-2$ B. $x-2 y+2=0$ C. $2 x+y=4$ D. $2 x+y-4=0$ Solution: Given that $y=x(2-x)$ $\Rightarrow y=2 x-x^{2}$ Slope of the tangent $\frac{d y}{d x}=2-2 x$ Slope at $(2,0)=2-4=-2$ Equation of normal: $\left(y-y_{1}\right)=\frac{-1}{\text { Slope of tangent }}\left(x-x_{1}\right)$ $\Rightarrow(y-0)=\frac{-1}{-2}(x-2)$ $\Rightarrow 2 y=x-2$ $\Rightarrow x-2 y-2=0$ Hence option $A$ is correct....
Read More →Following data is given for the reaction:
Question: Following data is given for the reaction: CaCO3 (s) CaO3 (s) + CO2 (g) ΔfH⊖ [CaO (s)] = -635.1 kJ mol=-1 ΔfH⊖ [CO2 (g)] = -393.5 kJ mol-1 ΔfH⊖ [CaCO3 (s)] = -1206.9 kJ mol-1 Predict the effect of temperature on the Equilibrium constant of the above reaction. Solution: ΔfH⊖ = ΔfH⊖products- ΔfH⊖reactants ΔfH⊖ = ΔfH⊖ [CaO(s)] + ΔfH⊖ [CO2(g)]- ΔfH⊖ [CaCO3(s)] = -635.1 -393.5+1206.9 = 178.3 kJ mol-1 =positive i.e. the reaction is exothermic. Hence according to Le Chateliers Principle on inc...
Read More →Using slopes, find the value of x for which the points A
Question: Using slopes, find the value of x for which the points A(5, 1), B(1, -1) and C(x, 4) are collinear. Solution: For three points to be collinear, the slope of all pairs must be equal, that is the slope of AB = slope of BC = slope of CA Given points are A(5, 1), B(1, -1) and C(x, 4) slope $=\left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)$ Slope of $A B=\left(\frac{-1-1}{1-5}\right)=\frac{-2}{-4}=\frac{1}{2}=0.5$ The slope of $B C=\left(\frac{4+1}{x-1}\right)=\left(\frac{5}{x-1}\right)$ Slope...
Read More →A reaction between ammonia and boron trifluoride
Question: A reaction between ammonia and boron trifluoride is given below: : NH3 + BF3 H3N: BF3 Identify the acid and base in this reaction. Which theory explains it? What is the hybridisation of B and N in the reactants? Solution: Lewis acid in this reaction is NH3 (N=1s22s22p1)as it has a lone pair of e- to donate in its p-orbital and Lewis base Is BF3 as p-orbital of Boron is empty(B=1s22s22p1) so it will accept lone pair from N and form a dative bond. This is explained by lewis electronic th...
Read More →Calculate the volume of water required
Question: Calculate the volume of water required to dissolve 0.1 g lead (II) chloride to get a saturated solution. (Ksp of PbCl2 = 3.2 10-8, atomic mass of Pb = 207 u). Solution: Given, Ksp of PbCl2 =3.2 10-8 The equation of disassociation of PbCl2 will be- PbCl2 ⇌ Pb2+ + 2Cl- Ksp = [Pb2+] [Cl-] 2 = (x) (2x) 2 = 43 43 =3.2 10-8 x=2 10-3 mol/L Solubility = molar mass (PbCl2) 2 10-3 =556 10-3 =0.556 g/L 0.1g of PbCl2 will dissolve in 0.1/0.0556 = 0.1798L The required volume to get a saturated solu...
Read More →The equation of the normal to the curve
Question: The equation of the normal to the curve A. $x=2$ B. $x=\pi$ C. $x+\pi=0$ D. $2 x=\pi$ Solution: Given that the curve $y=x+\sin x \cos x$ Differentiating both the sides w.r.t. $x$, $\frac{d y}{d x}=1+\cos ^{2} x-\sin ^{2} x$ Now, Slope of the tangent $\frac{\mathrm{dy}}{\mathrm{dx}}\left(\mathrm{x}=\frac{\pi}{2}\right)=1+\cos ^{2} \frac{\pi}{2}-\sin ^{2} \frac{\pi}{2}$ $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=1-1+0=0$ When $\mathrm{x}=\frac{\pi}{2}, \mathrm{y}=\frac{\pi}{2}$ Equatio...
Read More →The solubility product of Al(OH) 3 is 2.7 × 10-11.
Question: The solubility product of Al(OH) 3 is 2.7 10-11. Calculate its solubility in gL-1 and also find out the pH of this solution. (Atomic mass of Al = 27 u). Solution: The equation of disassociation of Al(OH)3will be- Al(OH)3 ⇌ Al3+ + 3OH- We know that, Ksp= [Al3+] [OH-] 3- =(s) (3s) 3 =27s4 S4 = Ksp/27 = 2.7x 10-11/27 s4= 10-12 s= (10-12)1/4 =10-3 mol/L Now, molar mass of Al(OH)3 =78 Solubility= molar mass s = 78 10-3 = 7.8 10-2 g/L NOW, we know that- pH = 14 pOH [OH]= 3s = 3 10-3 pOH= 3-l...
Read More →Using slopes show that the points A
Question: Using slopes show that the points A(6, -1), B(5, 0) and C(2, 3) are collinear. Solution: For three points to be collinear, the slope of all pairs must be equal, that is the slope of AB = slope of BC = slope of CA Given points are A(6, -1), B(5, 0) and C(2, 3) slope $=\left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)$ Slope of $A B=\left(\frac{0+1}{5-6}\right)=\frac{1}{-1}=-1$ Slope of $B C=\left(\frac{3-0}{2-5}\right)=\frac{3}{-3}=-1$ Slope of $C A=\left(\frac{3+1}{2-6}\right)=\frac{4}{-4}=...
Read More →Calculate the pH of a solution formed
Question: Calculate the pH of a solution formed by mixing equal volumes of two solutions A and B of a strong acid having pH = 6 and pH = 4 respectively. Solution: Given, pH of solution A = 6 [H+] of solution A = 10-6 mol/lit. pH of solution B = 4 [H+] of solution B = 10-4 mol/lit. On mixing 1L of each solution we will get total 2L of Solution. Amount of [H+] in 1L: solution A = 10-6 1L = 10-6 : Solution B = 10-4 1L = 10-4 Total [H+] in Solution = 10-6 + 10-4/2 =10-4 (1+0.01/2)=10-4 1.01/2 =510-5...
Read More →Without using Pythagora’s theorem, show that the points A
Question: Without using Pythagoras theorem, show that the points A(1, 2), B(4, 5) and C(6, 3) are the vertices of a right-angled triangle. Solution: The ΔABC is made up of three lines, AB,BC and CA For a right angle triangle, two lines must be at $90^{\circ}$ so they are perpendicular to each other. Checking for lines AB and BC For two lines to be perpendicular, their product of slope must be equal to -1. Given points A(1, 2), B(4, 5) and C(6, 3) slope $=\left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\rig...
Read More →pH of 0.08 mol dm-3 HOCl
Question: pH of 0.08 mol dm-3 HOCl solution is 2.85. Calculate its ionisation constant. Solution: Given, pH =2.85 and C =0.08 mol dm-3 Now since HOCl is a weak acid its dissociation will be given as- HOCl + H2O ⇌ H30+ + OCl- We know that; pH = -log [H+] -2.85 = log [H+] [H+] = antilog (-2.85) [H+] = 1.41 10-3 We also know that, for a weak mono basic acid- (On squaring both sides) [H+]2 = kaC Ka = (1.41 x 10-3)2/ 0.08 = 2.5 10-5 Hence the ionization constant of HOCl will be 2.5 10-5...
Read More →A sparingly soluble salt gets precipitated
Question: A sparingly soluble salt gets precipitated only when the product of the concentration of its ions in the solution (Qsp) becomes greater than its solubility product. If the solubility of BaSO4 in water is 8 10-4 mol dm-3, calculate its solubility in 0.01 mol dm-3 of H2SO4. Solution: Given, the solubility of BaSO4 in water= 8 10-4 g/L The equation of disassociation of BaSO4 will be- BaSO4 ⇌ Ba2+ + SO42- (S is the solubility of Ba2+ in 0.01 HCl) S 0.01, so it can be neglected We know that...
Read More →The pH of a solution of a strong acid is 5.0.
Question: The pH of a solution of a strong acid is 5.0. What will be the pH of the solution obtained after diluting the given solution 100 times? Solution: As we dilute the solution, the concentration of the solution will reduce by how much times we are diluting it. When diluted 100 times [h+] = 10-5/100 = 10-7mol/L And, pH Value will be = pH = -log [H+] =-log [10-7] = 7 Hence the pH after diluting solution a hundred times will be 7....
Read More →Based on the equation pH = – log [H+],
Question: Based on the equation pH = log [H+], the pH of 10-8 mol dm-3 solution of HCl should be 8. However, it is observed to be less than 7.0. Explain the reason. Solution: The solution is very dilute here and we know that HCl reacts with water to form hydronium ion. A decrease in pH can be observed as a result of a large concentration of H+. Hydronium ion concentration also to be considered here. Now total pH will be; [H3O+] = 10-8 + 10-7 M = 7 Hence the solution will be acidic....
Read More →Solve this
Question: If A(2, -5), B(-2, 5), C(x, 3) and D(1, 1) be four points such that AB and CD are perpendicular to each other, find the value of x. Solution: For two lines to be perpendicular, their product of slope must be equal to -1. Given points are A(2, -5),B(-2, 5) and C(x, 3),D(1, 1) slope $=\left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)$ $\Rightarrow$ Slope of line $A B$ is equal to $\left(\frac{5+5}{-2-2}\right)$ $=\left(\frac{10}{-4}\right)$ $=\left(\frac{-5}{2}\right)$ $=-2.5$ And the slope o...
Read More →The value of Kc for the reaction
Question: The value of Kc for the reaction 2HI (g) ⇋ H2 (g) + I2 (g) is 1 10-4 At a given time, the composition of the reaction mixture is [HI] =2 10-5 mol, [H2] =1 10-5 mol and [I2] =1 10-5 mol. In which direction will the reaction proceed? Solution: Given that Kc = 110-4 Kc = [H2][I2]/[HI]2 Qc expresses the relative ratio of products to reactants at a given instant. Qc = [H2][I2]/[HI]2 = (110-5)(110-5)/(210-4) Qc = 1/4 = 0.25 Here; QC KC Reaction will proceed in reverse direction....
Read More →Arrange the following in increasing
Question: Arrange the following in increasing order of pH KNO3 (aq), CH3COONa (aq), NH4Cl (aq), C6 H5COONH4 (aq) Solution: The increasing order of pH will be; CH3COONa KNO3 C6H5COONH4NH4Cl CH3COONa is a salt of a weak acid (CH3COOH) and strong base (NaOH) KNO3 is a salt of strong acid (HNO3)-strong base (KOH) C6H5COONH4 is a salt of a weak acid (benzoic acid) and weak base (NH4OH) NH4Cl is a salt of a strong acid (HCl) and a weak base (NH4OH)...
Read More →Show that the line through the points
Question: line through the points (-2, 6) and (4, 8) is perpendicular to the line through the points (3, -3) and (5, -9). Solution: For two lines to be perpendicular, their product of slope must be equal to -1. Given points are A(-2,6),B(4,8) and C(3,-3),D(5,-9) slope $=\left(\frac{\mathrm{y}_{2}-\mathrm{y}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}\right)$ Slope of line ABslope of line CD = -1 $\Rightarrow\left(\frac{8-6}{4+2}\right) \times\left(\frac{-9+3}{5-3}\right)=-1$ $\Rightarrow\left(\frac{2}{6...
Read More →The conjugate acid of a weak base is always
Question: The conjugate acid of a weak base is always stronger. What will be the decreasing order of basic strength of the following conjugate bases? OH-, RO-, CH3COO-, Cl- Solution: The conjugate base of a strong acid is weak therefore the decreasing order of basic strength will be; RO- OH- CH3COO- Cl-...
Read More →Ionization constant of a weak base
Question: Ionization constant of a weak base MOH is given by the expression Kb = [M+][OH-]/[MOH-] Values of ionisation constant of some weak bases at a particular temperature are given below: Base Di-methylamine Urea Pyridine Ammonia Kb 5.4 10-4 1.3 10-14 1.77 10-9 1.77 10-5 Arrange the bases in decreasing order of the extent of their ionisation at Equilibrium. Which of the above base is the strongest? Solution: The decreasing order of bases based on the ionisation constant at Equilibrium will b...
Read More →BF3 does not have proton but still
Question: BF3 does not have proton but still acts as an acid and reacts with NH3. Why is it so? What type of bond is formed between the two? Solution: According to Lewis concept e- deficient species are called lewis acid so BF3 will act as a lewis acid while NH3 (N=1s2 2s2 2p3) has a lone pair so it will act as a lewis base and it will donate its lone pair to the empty p-orbital of Boron through a coordinate bond to form an adduct....
Read More →Find the value of x so that the line through
Question: Find the value of x so that the line through (3, x) and (2, 7) is parallel to the line through (-1, 4) and (0, 6). Solution: We know that for two lines to be parallel, their slope must be the same. The given points are A(3,x),B(2,7) and C(-1,4),D(0,6) slope $=\left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)$ $\Rightarrow\left(\frac{6-4}{0+1}\right)=\left(\frac{7-x}{2-3}\right)$ $\Rightarrow\left(\frac{2}{1}\right)=\left(\frac{7-x}{-1}\right) \Rightarrow-2=7-x$ ⇒x = 9...
Read More →The equation to the normal to the curve
Question: The equation to the normal to the curve $y=\sin x$ at $(0,0)$ is A. $x=0$ B. $y=0$ C. $x+y=0$ D. $x-y=0$ Solution: Given that $y=\sin x$ Slope of the tangent $\frac{d y}{d x}=\cos x$ Slope at origin $=\cos 0=1$ Equation of normal: $\left(y-y_{1}\right)=\frac{-1}{\text { Slope of tangent }}\left(x-x_{1}\right)$ $\Rightarrow(\mathrm{y}-0)=\frac{-1}{1}(\mathrm{x}-0)$ $\Rightarrow y+x=0$ Hence option C is correct....
Read More →The aqueous solution of sugar does not conduct
Question: The aqueous solution of sugar does not conduct electricity. However, when sodium chloride is added to water, it conducts electricity. How will you explain this statement on the basis of ionization and how is it affected by the concentration of sodium chloride? Solution: The aqueous solution of sugar does not conduct electricity because they exist as a molecule in water. They dont have free ions to conduct electricity but in the case of NaCl, free ions of Na+ and Cl- are present to cond...
Read More →The equation to the normal to the curve
Question: The equation to the normal to the curve $y=\sin x$ at $(0,0)$ is A. $x=0$ B. $y=0$ C. $x+y=0$ D. $x-y=0$ Solution: Given that $y=\sin x$ Slope of the tangent $\frac{d y}{d x}=\cos x$ Slope at origin $=\cos 0=1$ Equation of normal: $\left(y-y_{1}\right)=\frac{-1}{\text { Slope of tangent }}\left(x-x_{1}\right)$ $\Rightarrow(\mathrm{y}-0)=\frac{-1}{1}(\mathrm{x}-0)$ $\Rightarrow \mathrm{y}+\mathrm{x}=0$ $\Rightarrow y+x=0$ Hence option C is correct....
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