Very-Short and Short-Answer Questions
Question: Very-Short and Short-Answer QuestionsThe first three terms of an AP are respectively (3y 1),(3y+ 5) and (5y+ 1), find the value ofy. Solution: The terms (3y 1),(3y+ 5) and (5y+ 1) are in AP. $\therefore(3 y+5)-(3 y-1)=(5 y+1)-(3 y+5)$ $\Rightarrow 3 y+5-3 y+1=5 y+1-3 y-5$ $\Rightarrow 6=2 y-4$ $\Rightarrow 2 y=10$ $\Rightarrow y=5$ Hence, the value ofyis 5....
Read More →Consider the above reaction.
Question: Consider the above reaction. The percentage yield of amide product is __________________ (Round off to the Nearest Integer). (Given : Atomic mass: $\mathrm{C}: 12.0 \mathrm{u}, \mathrm{H}: 1.0 \mathrm{u}$ $\mathrm{N}: 14.0 \mathrm{u}, \mathrm{O}: 16.0 \mathrm{u}, \mathrm{Cl}: 35.5 \mathrm{u})$ Solution: (77) $\therefore$ Theoretical amount of given product formed $=\frac{273}{140.5} \times 0.140=0.272 \mathrm{gm}$ But its actual amount formed is $0.210 \mathrm{gm}$. Hence, the percenta...
Read More →A child puts one five-rupee coin of her savings in the piggy bank on the first day.
Question: A child puts one five-rupee coin of her savings in the piggy bank on the first day. She increases her saving by one five-rupee coin daily. If the piggy bank can hold 190 coins of five rupees in all, find the number of days she can contribute to put the five-rupee coins into it and find the total money she saved. Solution: Saving of the child on the first day = Rs 5Saving on the second day = Rs 5 + Rs 5 = Rs 10Saving on the third day = Rs 5 + 2 Rs 5 = Rs 15 and so onThe saving of the ch...
Read More →A contract on construction job specifies a penalty for delay of completion beyond a a certain date as follows:
Question: A contract on construction job specifies a penalty for delay of completion beyond a a certain date as follows: ₹200 for the first day, ₹250 for the second day, ₹300 for the third day, etc., the penalty for each succeeding day being ₹50 more than for the preceding day. How much money the contractor has to pay as penalty, if he delayed the work by 30 days? Solution: It is given that the penalty for each succeeding day is ₹50 more than for the preceding day, so the amount of penalties are...
Read More →Solve the following
Question: $\mathrm{KBr}$ is doped with $10^{-5}$ mole percent of $\mathrm{SrBr}_{2}$. The number of cationic vacancies in $1 \mathrm{~g}$ of $\mathrm{KBr}$ crystal is ___________________ $10^{14}$. (Round off to the Nearest Integer). [Atomic Mass : $\mathrm{K}: 39.1 \mathrm{u}, \mathrm{Br}: 79.9 \mathrm{u}$$\left.\mathrm{N}_{\mathrm{A}}=6.023 \times 10^{23}\right]$ Solution: (5) 1 mole $\mathrm{KBr}(=119 \mathrm{gm})$ have $\frac{10^{-5}}{100}$ moles $\mathrm{SrBr}_{2}$ and hence, $10^{-7}$ mole...
Read More →The mole fraction of a solute in a 100 molal aqueous solution
Question: The mole fraction of a solute in a 100 molal aqueous solution__________________ $\times 10^{-2}$ (Round off to the Nearest Integer). Solution: (64) 100 molal aqueous solution means there is 100 mole solute in $1 \mathrm{~kg}=1000 \mathrm{gm}$ water. Now, mole-fraction of $=\frac{\mathrm{n}_{\text {solute }}}{\mathrm{n}_{\text {solute }}+\mathrm{n}_{\text {solvent }}}$ $=\frac{100}{100+\frac{1000}{18}}=\frac{1800}{2800}=0.6428$ $=64.28 \times 10^{-2}$...
Read More →In the above reaction, 3.9 g of benzene on nitration gives 4.92g of nitrobenzene.
Question: In the above reaction, $3.9 \mathrm{~g}$ of benzene on nitration gives $4.92 \mathrm{~g}$ of nitrobenzene. The percentage yield of nitrobenzene in the above reaction is _________ %. (Round off to the Nearest Integer). Given atomic mass : $\mathrm{C}: 12.0 \mathrm{u}, \mathrm{H}: 1.0 \mathrm{uO}: 16.0 \mathrm{u}, \mathrm{N}: 14.0 \mathrm{u}$ Solution: (80) $\begin{array}{ll}1 \text { mole } 1 \text { mole } \\ 78 \mathrm{gm} 123 \mathrm{gm}\end{array}$ $3.9 \mathrm{gm} \quad \frac{123}{...
Read More →A man arranges to pay off a debt of ₹36000 by 40 monthly instalments which form an arithmetic series.
Question: A man arranges to pay off a debt of ₹36000 by 40 monthly instalments which form an arithmetic series. When 30 of the instalments are paid, he dies leaving one-third of the debt unpaid. Find the value of the first instalment. Solution: Let the value of the first instalment be ₹a.Since the monthly instalments form an arithmetic series, so let us suppose the man increases the value of each instalment by ₹devery month. Common difference of the arithmetic series = ₹d Amount paid in 30 insta...
Read More →A 6.50 molal solution of KOH (aq.)
Question: A $6.50$ molal solution of $\mathrm{KOH}$ (aq.) has a density of $1.89 \mathrm{~g} \mathrm{~cm}^{-3}$. The molarity of the solution is _________moldm $^{-3}$. (Round off to the Nearest Integer). Solution: (9) $6.5$ molal $\mathrm{KOH}=1000 \mathrm{gm}$ solvent has $6.5$ moles $\mathrm{KOH}$ so $\mathrm{w}$ t of solute $=6.5 \times 56$ $=364 \mathrm{gm}$ wt of solution $=1000+364=1364$ Volume of solution $=\frac{1364}{1.89} \mathrm{~m} \ell$ Molarity $=\frac{\text { mole of solute }}{\m...
Read More →A man saved ₹33000 in 10 months. In each month after the first, he saved ₹100 more than he did in the preceding month
Question: A man saved ₹33000 in 10 months. In each month after the first, he saved ₹100 more than he did in the preceding month. How much did he save in the first month? Solution: Let the money saved by the man in the first month be ₹a.It is given that in each month after the first, he saved ₹100 more than he did in the preceding month. So, the money saved by the man every month is in AP with common difference ₹100.d= ₹100Number of months,n= 10Sum of money saved in 10 months,S10= ₹33,000 Using t...
Read More →A sum of ₹700 is to be used to give seven cash prizes to students of a school for their overall academic performance.
Question: A sum of ₹700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹20 less than its preceding prize, find the value of each prize. Solution: Let the value of the first prize be ₹a.Since the value of each prize is ₹20 less than its preceding prize, so the values of the prizes are in AP with common difference ₹20.d= ₹20Number of cash prizes to be given to the students,n= 7Total sum of the prizes,S7= ₹700 Using the formu...
Read More →Complete combustion of 750g of an organic compound provides 420 g of
Question: Complete combustion of $750 \mathrm{~g}$ of an organic compound provides $420 \mathrm{~g}$ of $\mathrm{CO}_{2}$ and $210 \mathrm{~g}$ of $\mathrm{H}_{2} \mathrm{O}$. The percentage composition of carbon and hydrogen in organic compound is $15.3$ and....respectively. (Round off to the Nearest Integer) Solution: (3) $44 \mathrm{gm} \mathrm{CO}_{2}$, have $12 \mathrm{gm}$ carbon So, $420 \mathrm{gmCO}_{2} \Rightarrow \frac{12}{44} \times 420$ $\Rightarrow \frac{1260}{11}$ gm carbon $\Righ...
Read More →There are 25 trees at equal distances of 5 m in a line with a water tank, the distance of the water tank from the nearest tree being 10 m
Question: There are 25 trees at equal distances of 5 m in a line with a water tank, the distance of the water tank from the nearest tree being 10 m. A gardener waters all the trees separately, starting from the water tank and returning back to the water tank after watering each tree to get water for the next. Find the total distance covered by the gardener in order to water all the trees. Solution: Distance covered by the gardener to water the first tree and return to the water tank = 10 m + 10 ...
Read More →In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line.
Question: In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are 10 potatoes in the line. A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and he continues in the same way until all the potatoes are in the bucket. What is the total distance the competit...
Read More →In a school, students decided to plant trees in and around the school to reduce air pollution.
Question: In a school, students decided to plant trees in and around the school to reduce air pollution. It was decided that the number of trees that each section of each class will plant will be double of the class in which they are studying. If there are 1 to 12 classes in the school and each class has two sections, find how many trees were planted by students. Which value is shown in the question? Solution: Number of trees planted by the students of each section of class 1 = 2There are two se...
Read More →Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Question: Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively. Solution: Letabe the first term anddbe the common difference of the AP. Then, $d=a_{3}-a_{2}=18-14=4$ Now, $a_{2}=14$ (Given) $\Rightarrow a+d=14 \quad\left[a_{n}=a+(n-1) d\right]$ $\Rightarrow a+4=14$ $\Rightarrow a=14-4=10$ Using the formula, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$, we get $S_{51}=\frac{51}{2}[2 \times 10+(51-1) \times 4]$ $=\frac{51}{2}(20+200)$ $=\frac{51}{2} \times 220$ $=561...
Read More →How much amount of NaCl should be added to 600 g of water
Question: How much amount of $\mathrm{NaCl}$ should be added to $600 \mathrm{~g}$ of water $(\rho=1.00 \mathrm{~g} / \mathrm{mL})$ to decrease the freezing point of water to $-0.2^{\circ} \mathrm{C}$ ? _________________ . (The freezing point depression constant for water $=2 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ ) Solution: (1.75) $\Delta \mathrm{T}_{\mathrm{f}}=i k_{\mathrm{f}} m$ $0.2=2 \times 2 \times \frac{w}{58.5} \times \frac{1000}{600}$ $w=\frac{0.2 \times 58.5 \times 600}{1000 \ti...
Read More →Sum of the first 14 terms of an AP is 1505 and its first term is 10. Find its 25th term.
Question: Sum of the first 14 terms of an AP is 1505 and its first term is 10. Find its 25th term. Solution: Letdbe the common difference of the AP.Here,a= 10 andn= 14Now, $S_{14}=1505$ (Given) $\Rightarrow \frac{14}{2}[2 \times 10+(14-1) \times d]=1505 \quad\left\{S_{n}=\frac{n}{2}[2 a+(n-1) d]\right\}$ $\Rightarrow 7(20+13 d)=1505$ $\Rightarrow 20+13 d=215$ $\Rightarrow 13 d=215-20=195$ $\Rightarrow d=15$ 25th term of the AP,a25 $=10+(25-1) \times 15 \quad\left[a_{n}=a+(n-1) d\right]$ $=10+360...
Read More →'X' melts at low temperature and is a bad conductor of
Question: 'X' melts at low temperature and is a bad conductor of electricity in both liquid and solid state. $X$ is:Zinc sulphideMercurySilicon carbideCarbon tetrachlorideCorrect Option: , 4 Solution: $\mathrm{CCl}_{4} \rightarrow$ Non-conductor in solid and liquid phase. Melting point of $\mathrm{CCl}_{4}$ is $-23{ }^{\circ} \mathrm{C}$. It does not conduct electricity in both solid and liquid state....
Read More →Find the number of terms of the AP −12, −9, −6, ..., 21. If 1 is added to each term of this AP then find the sum of all terms of the
Question: Find the number of terms of the AP 12, 9, 6, ..., 21. If 1 is added to each term of this AP then find the sum of all terms of the AP thus obtained. Solution: The given AP is 12, 9, 6, ..., 21.Here,a= 12,d= 9 (12) =9 + 12 = 3 andl= 21Suppose there arenterms in the AP. $\therefore l=a_{n}=21$ $\Rightarrow-12+(n-1) \times 3=21 \quad\left[a_{n}=a+(n-1) d\right]$ $\Rightarrow 3 n-15=21$ $\Rightarrow 3 n=21+15=36$ $\Rightarrow n=12$ Thus, there are 12 terms in the AP.If 1 is added to each te...
Read More →Two open beakers one containing a solvent and the other containing a mixture
Question: Two open beakers one containing a solvent and the other containing a mixture of that solvent with a non volatile solute are together sealed in a container. Over time:the volume of the solution increases and the volume of the solvent decreasesthe volume of the solution decreases and the volume of the solvent increasesthe volume of the solution and the solvent does not changethe volume of the solution does not change and the volume of the solvent decreasesCorrect Option: 1 Solution: Ther...
Read More →Solve the following
Question: At $35^{\circ} \mathrm{C}$, the vapour pressure of $\mathrm{CS}_{2}$ is $512 \mathrm{~mm} \mathrm{Hg}$ and that of acetone is $344 \mathrm{~mm} \mathrm{Hg}$. A solution of $\mathrm{CS}_{2}$ in acetone has a total vapour pressure of $600 \mathrm{~mm} \mathrm{Hg}$. The false statement amongst the following is:Raoult's law is not obeyed by this systema mixture of $100 \mathrm{~mL} \mathrm{CS}_{2}$ and $100 \mathrm{~mL}$ acetone has a volume $200 \mathrm{~mL}$$\mathrm{CS}_{2}$ and acetone ...
Read More →The sum of first q terms of an AP is (63q − 3q2). If its pth term is −60, find the value of p.
Question: The sum of firstqterms of an AP is (63q 3q2). If itspth term is 60, find the value ofp. Also, find the 11th term of its AP. Solution: LetSqdenote the sum of the firstqterms of the AP. Then, $S_{q}=63 q-3 q^{2}$ $\Rightarrow S_{q-1}=63(q-1)-3(q-1)^{2}$ $=63 q-63-3\left(q^{2}-2 q+1\right)$ $=-3 q^{2}+69 q-66$ Supposeaqdenote theqthterm of the AP. $\therefore a_{q}=S_{q}-S_{q-1}$ $=\left(63 q-3 q^{2}\right)-\left(-3 q^{2}+69 q-66\right)$ $=-6 q+66 \quad \ldots(1)$ Now, $a_{p}=-60$ (Given)...
Read More →A set of solutions is prepared using 180 g of water as a solvent and 10g of different non-volatile solutes A, B and C.
Question: A set of solutions is prepared using $180 \mathrm{~g}$ of water as a solvent and $10 \mathrm{~g}$ of different non-volatile solutes A, B and C. The relative lowering of vapour pressure in the presence of these solutes are in the order [Given, molar mass of $\mathrm{A}=100 \mathrm{~g} \mathrm{~mol}^{-1} ; \mathrm{B}=200 \mathrm{~g} \mathrm{~mol}^{-1} ; \mathrm{C}=10,000 \mathrm{~g} \mathrm{~mol}^{-1}$ ]$\mathrm{B}\mathrm{C}\mathrm{A}$$\mathrm{C}\mathrm{B}\mathrm{A}$$\mathrm{A}\mathrm{B}...
Read More →The sum of first m terms of an AP is (4m2 − m). If its nth term is 107, find the value of n.
Question: The sum of firstmterms of an AP is (4m2m). If itsnth term is 107, find the value ofn. Also, find the 21st term of this AP. Solution: LetSmdenote the sum of the firstmterms of the AP. Then, $S_{m}=4 m^{2}-m$ $\Rightarrow S_{m-1}=4(m-1)^{2}-(m-1)$ $=4\left(m^{2}-2 m+1\right)-(m-1)$ $=4 m^{2}-9 m+5$ Supposeamdenote themthterm of the AP. $\therefore a_{m}=S_{m}-S_{m-1}$ $=\left(4 m^{2}-m\right)-\left(4 m^{2}-9 m+5\right)$ $=8 m-5 \quad \ldots(1)$ Now, $a_{n}=107$ (Given) $\Rightarrow 8 n-5...
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