Among the isomeric alkanes of molecular formula

Question: Among the isomeric alkanes of molecular formula C5H12, identify the one that on photochemical chlorination yields (i)A single monochloride. (ii)Three isomeric monochlorides. (iii)Four isomeric monochlorides. Solution: (i)To have a single monochloride, there should be only one type of H-atom in the isomer of the alkane of the molecular formula C5H12. This is because, replacement of any H-atom leads to the formation of the same product. The isomer is neopentane. Neopentane (ii)To have th...

Read More →

Write structures of different dihalogen derivatives of propane.

Question: Write structures of different dihalogen derivatives of propane. Solution: There are four different dihalogen derivatives of propane. The structures of these derivatives are shown below. (i) 1, 1-Dibromopropane (ii) 2, 2-Dibromopropane (iii) 1, 2-Dibromopropane (iv) 1, 3-Dibromopropane...

Read More →

∆ABC is an isosceles triangle with AB = AC = 13 cm and the length of altitude from

Question: ∆ABCis an isosceles triangle withAB=AC= 13 cm and the length of altitude fromAonBCis 5 cm. Then,BC= ?(a) 12 cm(b) 16 cm(c) 18 cm(d) 24 cm Solution: (d) 24 cm In triangle ABC, let the altitude fromAonBCmeetsBCatD.We have:AD= 5 cm,AB= 13 cm andDis the midpoint ofBC.Applying Pythagoras theorem in right-angled triangleABD,we get: $A B^{2}=A D^{2}+B D^{2}$ $\Rightarrow B D^{2}=A B^{2}-A D^{2}$ $\Rightarrow B D^{2}=13^{2}-5^{2}$ $\Rightarrow B D^{2}=169-25$ $\Rightarrow B D^{2}=144$ $\Righta...

Read More →

Why is sulphuric acid not used during the reaction of alcohols with KI?

Question: Why is sulphuric acid not used during the reaction of alcohols with KI? Solution: In the presence of sulphuric acid (H2SO4), KI produces HI $2 \mathrm{KI}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow 2 \mathrm{KHSO}_{4}+2 \mathrm{HI}$ Since $\mathrm{H}_{2} \mathrm{SO}_{4}$ is an oxidizing agent, it oxidizes $\mathrm{HI}$ (produced in the reaction to $\mathrm{I}_{2}$ ). $2 \mathrm{HI}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{I}_{2}+\mathrm{SO}_{2}+\mathrm{H}_{2} \mathrm{...

Read More →

A conical hole is drilled in a circular cylinder of height 12 cm

Question: A conical hole is drilled in a circular cylinder of height 12 cm and base radius 5 cm. The height and the base radius of the cone are also the same. Find the whole surface and volume of the remaining cylinder. Solution: Given that: $r=5 \mathrm{~cm}$ $h=12 \mathrm{~cm}$ We have the following diagram Slant height of cone is given by $l=\sqrt{r^{2}+h^{2}}$ $=\sqrt{5^{2}+12^{2}}$ $=13 \mathrm{~cm}$ The total surface area of the remaining part is given by $S=2 \pi r h+\pi r^{2}+\pi r l$ $=...

Read More →

Write structures of the following compounds:

Question: Write structures of the following compounds: (i)2-Chloro-3-methylpentane (ii)1-Chloro-4-ethylcyclohexane (iii)4-tert. Butyl-3-iodoheptane (iv)1,4-Dibromobut-2-ene (v)1-Bromo-4-sec. butyl-2-methylbenzene Solution: (i) 2-Chloro-3-methyl pentane (ii) 1-Chloro-4-ethylcyclohexane (iii) 4- tert-Butyl-3-iodoheptane (iv) 1, 4-Dibromobut-2-ene (v) 1-Bromo-4-sec-butyl-2-methylbenzene...

Read More →

A petrol tank is a cylinder of base diameter 21 cm and length 18 cm

Question: A petrol tank is a cylinder of base diameter 21 cm and length 18 cm fitted with conical ends each of axis length 9 cm. Determine the capacity of the tank. Solution: To find the total capacity of the tank, we have to add the volume of the cylinder and cone. Diameter of the cylinder, $d=21 \mathrm{~cm}$ Radius of the cylinder, $r=\frac{d}{2}=\frac{21}{2} \mathrm{~cm}$ Height of the cylinder, $h_{1}=18 \mathrm{~cm}$ Also, radius of cone, $r=\frac{21}{2} \mathrm{~cm}$ Also, radius of cone,...

Read More →

The height of an equilateral triangle having each side 12 cm, is

Question: The height of an equilateral triangle having each side 12 cm, is (a) $6 \sqrt{2} \mathrm{~cm}$ (b) $6 \sqrt{3} \mathrm{~cm}$ (c) $3 \sqrt{6} \mathrm{~cm}$ (d) $6 \sqrt{6} \mathrm{~cm}$ Solution: (b) $6 \sqrt{3} \mathrm{~cm}$ Let ABC be the equilateral triangle with AD as its altitude from A.In right-angled triangle ABD, we have: $A B^{2}=A D^{2}+B D^{2}$ $A D^{2}=A B^{2}-B D^{2}$ $=12^{2}-6^{2}$ $=144-36=108$ $A D=\sqrt{108}=6 \sqrt{3} \mathrm{~cm}$...

Read More →

The hypotenuse of a right triangle is 25 cm. The other two sides are such that one is 5 cm longer.

Question: The hypotenuse of a right triangle is 25 cm. The other two sides are such that one is 5 cm longer. The lengths of these sides are(a) 10 cm, 15 cm(b) 15 cm, 20 cm(c) 12 cm, 17 cm(d)13 cm, 18 cm Solution: (b) 15 cm, 20 cmIt is given that the length of hypotenuse is 25 cm. Let the other two sides be $x \mathrm{~cm}$ and $(x-5) \mathrm{cm}$. Applying Pythagoras theorem, we get: $25^{2}=x^{2}+(x-5)^{2}$ $\Rightarrow 625=x^{2}+x^{2}+25-10 x$ $\Rightarrow 2 x^{2}-10 x-600=0$ $\Rightarrow x^{2...

Read More →

What will be the correct order for the wavelengths of absorption in the visible region for the following:

Question: What will be the correct order for the wavelengths of absorption in the visible region for the following: [Ni(NO2)6]4, [Ni(NH3)6]2+, [Ni(H2O)6]2+ Solution: The central metal ion in all the three complexes is the same. Therefore, absorption in the visible region depends on the ligands. The order in which the CFSE values of the ligands increases in the spectrochemical series is as follows: H2O NH3 NO2 Thus, the amount of crystal-field splitting observed will be in the following order: He...

Read More →

Amongst the following, the most stable complex is

Question: Amongst the following, the most stable complex is (i)[Fe(H2O)6]3+ (ii)[Fe(NH3)6]3+ (iii)[Fe(C2O4)3]3 (iv)[FeCl6]3 Solution: We know that the stability of a complex increases by chelation. Therefore, the most stable complex is [Fe(C2O4)3]3....

Read More →

The oxidation number of cobalt in

Question: The oxidation number of cobalt in K[Co(CO)4] is (i)+1 (ii)+3 (iii)1 (iv)3 Solution: We know that CO is a neutral ligand and K carries a charge of +1. Therefore, the complex can be written as K+[Co(CO)4]. Therefore, the oxidation number of Co in the given complex is 1. Hence, option (iii) is correct....

Read More →

Amongst the following ions which one has the highest magnetic moment value?

Question: Amongst the following ions which one has the highest magnetic moment value? (i)[Cr(H2O)6]3+ (ii)[Fe(H2O)6]2+ (iii)[Zn(H2O)6]2+ Solution: (i) No. of unpaired electrons in $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}=3$ Then, $\mu=\sqrt{n(n+2)}$ $=\sqrt{3(3+2)}$ $=\sqrt{15}$ $\sim 4 \mathrm{BM}$ (ii)No. of unpaired electrons in [Fe(H2O)6]2+= 4 Then, $\mu=\sqrt{4(4+2)}$ $=\sqrt{24}$ $\sim 5 \mathrm{BM}$ (iii)No. of unpaired electrons in [Zn(H2O)6]2+= 0 Hence, [...

Read More →

In the given figure, O is the point inside a △MNP such that ∠MOP = 900 ,

Question: In the given figure, O is the point inside a △MNP such that MOP = 900, OM = 16 cm and OP = 12 cm. If MN = 21 cm and NMP = 900thenNP = ?(a) 25 cm(b) 29 cm(c) 33 cm(d) 35 cm Solution: Now, In right triangle MOPBy using Pythagoras theorem, we have $\mathrm{MP}^{2}=\mathrm{PO}^{2}+\mathrm{OM}^{2}$ $=12^{2}+16^{2}$ $=144+256$ $=400$ $\therefore \mathrm{MP}^{2}=400$ $\Rightarrow \mathrm{MP}=20 \mathrm{~cm}$ Now, In right triangle MPNBy using Pythagoras theorem, we have $\mathrm{PN}^{2}=\math...

Read More →

How many ions are produced from the complex

Question: How many ions are produced from the complex Co(NH3)6Cl2in solution? (i)6 (ii)4 (iii)3 (iv)2 Solution: (iii)The given complex can be written as [Co(NH3)6]Cl2. Thus,[Co(NH3)6]+along with two Clions are produced....

Read More →

Discuss briefly giving an example in each case the role of coordination compounds in:

Question: Discuss briefly giving an example in each case the role of coordination compounds in: (i) biological system (ii) medicinal chemistry (iii) analytical chemistry (iv) extraction/metallurgy of metals Solution: (i)Role of coordination compounds in biological systems: We know that photosynthesis is made possible by the presence of the chlorophyll pigment. This pigment is a coordination compound of magnesium. In the human biological system, several coordination compounds play important roles...

Read More →

What is meant by the chelate effect?

Question: What is meant by thechelate effect? Give an example. Solution: When a ligand attaches to the metal ion in a manner that forms a ring, then the metal- ligand association is found to be more stable. In other words, we can say that complexes containing chelate rings are more stable than complexes without rings. This is known as the chelate effect. For example:...

Read More →

A ladder 25 m long just reaches the top of a building 24 m high from the ground.

Question: A ladder 25 m long just reaches the top of a building 24 m high from the ground. What is the distance of the foot of the ladder from the building?(a) 7 m(b) 14 m(c) 21 m(d) 24.5 m Solution: (a) 7 m Let the ladder BC reaches the building at C.Let the height of building where the ladder reaches be AC.According to the question:BC = 25 mAC = 24 mIn right-angled triangle CAB, we apply Pythagoras theorem to find the value of AB. $B C^{2}=A C^{2}+A B^{2}$ $\Rightarrow A B^{2}=B C^{2}-A C^{2}=...

Read More →

What is meant by stability of a coordination compound in solution?

Question: What is meant by stability of a coordination compound in solution? State the factors which govern stability of complexes. Solution: The stability of a complex in a solution refers to the degree of association between the two species involved in a state of equilibrium. Stability can be expressed quantitatively in terms of stability constant or formation constant. For this reaction, the greater the value of the stability constant, the greater is the proportion of ML3in the solution. Stab...

Read More →

A cylindrical tub of radius 5 cm and length 9.8 cm

Question: A cylindrical tub of radius 5 cm and length 9.8 cm is full of water. A solid in the form of a right circular cone mounted on a hemisphere is immersed in the tub. If the radius of the hemisphere is immersed in the tub. If the radius of the hemi-sphere is 3.5 cm and height of the cone outside the hemisphere is 5 cm, find the volume of the water left in the tub (Take = 22/7) Solution: To find the volume of the water left in the tube, we have to subtract the volume of the hemisphere and co...

Read More →

The shadow of a 5-m-long stick is 2 m long. At the same time the length of the shadow of a 12.5 m

Question: The shadow of a 5-m-long stick is 2 m long. At the same time the length of the shadow of a 12.5 m high tree (in m) is(a) 3.0(b) 3.5(c) 4.5(d) 5.0 Solution: Suppose DE is a 5 m long stick and BC is a 12.5 m high tree.Suppsose DA and BA are the shadows of DE and BC respectively.Now, In △ABC and △ADEABC= ADE = 900 A = A (Common)By AA-similarity criterion△ABC △ADEIf two triangles are similar, then the the ratio of their corresponding sides are equal. $\therefore \frac{\mathrm{AB}}{\mathrm{...

Read More →

A toy is in the shape of a right circular cylinder

Question: A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The radius and height of the cylindrical part are 5 cm and 13 cm respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Find the surface area of the toy if the total height of the toy is 30 cm. Solution: We have the following diagram For cylindrical part, we have $h=13 \mathrm{~cm}$ $r=5 \mathrm{~cm}$ Therefore, the curved surfac...

Read More →

A vertical pole 6 m long casts a shadow of length 3.6 m on the ground.

Question: A vertical pole 6 m long casts a shadow of length 3.6 m on the ground. What is the height of a tower which casts a shadow of length 18 m at the same time?(a) 10.8 m(b) 28.8 m(c) 32.4 m(d) 30 m Solution: (d) 30 m Let AB and AC be the vertical pole and its shadow, respectively.According to the question:AB = 6 mAC = 3.6 mAgain, let DE and DFbe the tower and its shadow.According to the question:DF = 18 m​DE = ?Now, in right-angled triangles ABC and DEF, we have: $\angle B A C=\angle E D F=...

Read More →

A solid is in the form of a right circular cylinder,

Question: A solid is in the form of a right circular cylinder, with a hemisphere at one end and a cone at the other end. The radius of the common base is 3.5 cm and the heights of the cylindrical and conical portions are 10 cm. and 6 cm, respectively. Find the total surface area of the solid. (Use = 22/7) Solution: We have the following diagram For cone, we have $r=3.5 \mathrm{~cm}$ $h=6 \mathrm{~cm}$ $l=\sqrt{r^{2}+h^{2}}$ $=\sqrt{3.5^{2}+6^{2}}$ $=6.95 \mathrm{~cm}$ Curved surface area of the ...

Read More →

A vertical stick 1.8 m long casts a shadow 45 cm long on the ground. At the same time, what is the length of the shadow of a pole 6 m high?

Question: A vertical stick 1.8 m long casts a shadow 45 cm long on the ground. At the same time, what is the length of the shadow of a pole 6 m high?(a) 2.4 m(b) 1.35 m(c) 1.5 m(d) 13.5 m Solution: (c) 1.5 m Let AB and AC be the vertical stick and its shadow, respectively.According to the question:AB = 1.8mAC = 45 cm = 0.45 mAgain, let DE and DF be the pole and its shadow, respectively.According to the question:DE = 6 mDF = ?Now, in right-angled triangles ABC and DEF, we have: $\angle B A C=\ang...

Read More →