Some Basic Concepts of Chemistry
$20 \mathrm{~mL}$ of $0.02 \,\mathrm{M} \,\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ solution is used for the titration of $10 \mathrm{~mL}$ of $\mathrm{Fe}^{2+}$ solution in the acidic medium.
The molarity of $\mathrm{Fe}^{2+}$ solution is __________ $\times \,10^{-2}\, \mathrm{M}$. (Nearest Integer)
Explanation:
$ \begin{aligned} &10 \times 1 \times M=20 \times 6 \times .02 \\\\ &M=24 \times 10^{-2} M \end{aligned} $
$\therefore $ Answer will be 24
A $100 \mathrm{~mL}$ solution of $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{MgBr}$ on treatment with methanol produces $2.24 \mathrm{~mL}$ of a gas at STP. The weight of gas produced is _____________ mg. [nearest integer]
Explanation:
As $2.24 \,\mathrm{ml}$ is formed at STP.
Number of moles of ethane gas produced
$ =\frac{2.24 X}{22.4} $
$=10^{-4}\, \mathrm{ml}$
Mass of ethane produced $=10^{-4} \times 30=3 \times 10^{-}=3\mathrm{mg}$
Chlorophyll extracted from the crushed green leaves was dissolved in water to make $2 \mathrm{~L}$ solution of Mg of concentration $48\, \mathrm{ppm}$. The number of atoms of $\mathrm{Mg}$ in this solution is $x \times 10^{20}$ atoms. The value of $x$ is ___________. (Nearest Integer)
(Given : Atomic mass of $\mathrm{Mg}$ is $24 \mathrm{~g} \mathrm{~mol}^{-1} ; \mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1}$ )
Explanation:
Number of atoms of $\mathrm{Mg}=\frac{96 \times 10^{-3}}{24} \times \mathrm{N}_{\mathrm{A}}$
$ \begin{aligned} &=4 \times 10^{-3} \times 6 \times 10^{23} \\ &=24 \times 10^{20} \end{aligned} $
$x=24$
When 800 mL of 0.5 M nitric acid is heated in a beaker, its volume is reduced to half and 11.5 g of nitric acid is evaporated. The molarity of the remaining nitric acid solution is x $\times$ 10$-$2 M. (Nearest integer)
(Molar mass of nitric acid is 63 g mol$-$1)
Explanation:
Moles of $\mathrm{HNO}_{3}=400 \times 10^{-3} = 0.4 \text { moles }$
Weight of $\mathrm{HNO}_{3}=0.4 \times 63 \mathrm{~g} =25.2 \mathrm{~g} $
Remaining acid $=25.2-11.5 =13.7 \mathrm{~g} $
$ \begin{aligned} M &=\frac{13.7 \times 1000}{400 \times 63} \\ &=\frac{137}{252}=0.54 \\ &=54 \times 10^{-2} \end{aligned} $
56.0 L of nitrogen gas is mixed with excess hydrogen gas and it is found that 20 L of ammonia gas is produced. The volume of unused nitrogen gas is found to be _________ L.
Explanation:
Since $\mathrm{H}_{2}$ is in excess and $20 \mathrm{~L}$ of ammonia gas is produced.
Hence, 2 moles $\mathrm{NH}_{3} \equiv 1$ mole $\mathrm{N}_{2} \quad(v \propto \mathrm{n})$
$ 20 \mathrm{~L} \mathrm{NH}_{3} \equiv 10 \mathrm{~L} \mathrm{~N}_{2} $
Volume of $\mathrm{N}_{2}$ left $=56-10$
$ =46 \mathrm{~L} $
A sample of 4.5 mg of an unknown monohydric alcohol, R-OH was added to methylmagnesium iodide. A gas is evolved and is collected and its volume measured to be 3.1 mL. The molecular weight of the unknown alcohol is __________ g/mol. [Nearest integer]
Explanation:
moles of alcohol $(\mathrm{ROH}) \equiv$ moles of $\mathrm{CH}_{4}$
At STP, [Assuming STP]
1 mole corresponds to $22.7 \mathrm{~L}$
Hence, $3.1 \mathrm{~mL} \equiv \frac{3.1}{22700} \mathrm{~mol}$
So, moles of alcohol $=\frac{3.1}{22700}$
$\Rightarrow \frac{3.1}{22700}=\frac{4.5 \times 10^{-3}}{\mathrm{M}}$
$M \simeq 33 \mathrm{~g} / \mathrm{mol}$
Blister copper is produced by reaction of copper oxide with copper sulphide.
2Cu2O + Cu2S $\to$ 6Cu + SO2
When 2.86 $\times$ 103 g of Cu2O and 4.77 $\times$ 103 g of Cu2S are used for reaction, the mass of copper produced is _____________ g. (nearest integer)
(Atomic mass of Cu = 63.5 a.m. u, S = 32.0 a.m. u, O = 16.0 a.m. u)
Explanation:
moles of $\mathrm{Cu}_{2} \mathrm{S}=\frac{4.77 \times 10^{3}}{159}=30$
$\mathrm{Cu}_{2} \mathrm{O}$ is limiting reagent
$ \begin{array}{lccc} 2 \mathrm{Cu}_{2} \mathrm{O} & + & \mathrm{Cu}_{2} \mathrm{~S} \longrightarrow & 6 \mathrm{Cu} & + & \mathrm{SO}_{2} \\ 20 && 30 & - & & - \\ - & &20 & 60 & & 10 \end{array} $
Mass of copper $=60 \times 63.5=3810 \mathrm{~g}$
The complete combustion of 0.492 g of an organic compound containing 'C', 'H' and 'O' gives 0.793 g of CO2 and 0.442 g of H2O. The percentage of oxygen composition in the organic compound is ______________. (nearest integer)
Explanation:
Total organic compound = 0.492 gm
Produced CO2 = 0.793 gm
$\therefore$ Moles of CO2 = ${{0.793} \over {44}}$
$\therefore$ Moles of C atoms = ${{0.793} \over {44}}$
$\therefore$ Weight of C atoms = ${{0.793} \over {44}}$ $\times$ 12 = 0.216 g
Produced H2O = 0.442 gm
$\therefore$ Moles of H2O = ${{0.442} \over {18}}$
$\therefore$ Moles of H atoms = ${{0.442} \over {18}}$
$\therefore$ Weight of H atoms = ${{0.442} \over {18}}$ $\times$ 2 = 0.05 g
$\therefore$ Weight of O atoms
= 0.492 $-$ (0.216 + 0.05)
= 0.226 gm
% by mass of oxygen in compound
= ${{0.226} \over {0.492}}$ $\times$ 100 = 46%
116 g of a substance upon dissociation reaction, yields 7.5 g of hydrogen, 60 g of oxygen and 48.5 g of carbon. Given that the atomic masses of H, O and C are 1, 16 and 12, respectively. The data agrees with how many formulae of the following?
A. CH3COOH, B. HCHO, C. CH3OOCH3, D. CH3CHO
Explanation:
Relative atomicities $=\mathrm{H} \Rightarrow 6.5$
$ \begin{aligned} &\mathrm{O} \Rightarrow \frac{51.7}{16}=3.25 \\\\ &\mathrm{C} \Rightarrow \frac{41.8}{12}=3.5 \end{aligned} $
Emperically formula is approx.. $\mathrm{CH}_2 \mathrm{O}$
(A) $\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}_2$ (B) $\mathrm{CH}_2 \mathrm{O}$ relate to this formula.
Two elements A and B which form 0.15 moles of A2B and AB3 type compounds. If both A2B and AB3 weigh equally, then the atomic weight of A is _____________ times of atomic weight of B.
Explanation:
Let atomic weight of A = x
and atomic weight of B = y
$\therefore$ Molar mass of A2B = 2x + y
and molar mass of AB3 = x + 3y
Now, weight of 0.15 moles of A2B = (2x + y) 0.15
and weight of 0.15 moles of AB3 = (x + 3y) 0.15
Given, weight of 0.15 moles of A2B and AB3 are equal.
$\therefore$ (2x + y)0.15 = (x + 3y)0.15
$\Rightarrow$ x = 2y
$ \Rightarrow {x \over y} = 2$
$\therefore$ Atomic weight of A is 2 times of atomic weight of B.
CNG is an important transportation fuel. When 100 g CNG is mixed with 208 g oxygen in vehicles, it leads to the formation of CO2 and H2O and produces large quantity of heat during this combustion, then the amount of carbon dioxide, produced in grams is ____________. [nearest integer]
[Assume CNG to be methane]
Explanation:
Initially,
Mass of CNG(CH4) = 100 gm
$\therefore$ Moles of CH4 = ${{100} \over {16}}$ = 6.25 moles
Mass of O2 = 208 gm
$\therefore$ Moles of O2 = ${{208} \over {32}}$ = 6.5 moles
From reaction you can see,
1 mole of CH4 react with 2 mole of O2
$\therefore$ 6.25 mole of CH4 react with 6.25 $\times$ 2 = 12.5 moles of O2
But here only 6.5 moles of O2 present
So, O2 will act as limiting reagent.
$\therefore$ Produced CO2 will depends on moles of O2.
From 2 moles of O2 1 mole of CO2
$\therefore$ From 6.5 moles of O2 ${{6.5} \over {2}}$ moles of CO2 produced.
$\therefore$ Weight of CO2 = ${{6.5} \over {2}}$ $\times$ 44 = 143 gm
The moles of methane required to produce 81 g of water after complete combustion is _____________ $\times$ 10$-$2 mol. [nearest integer]
Explanation:
POAC on $\mathrm{H}$ atom
$ \begin{aligned} &\mathrm{n}_{\mathrm{CH} 4} \times 4=\mathrm{n}_{\mathrm{H} 2 \mathrm{O}} \times 2 \\\\ &\mathrm{n}_{\mathrm{CH}_4}=\frac{81}{18} \times 2 \times \frac{1}{4}=\frac{81}{36} \\\\ &\mathrm{n}_{\mathrm{CH}_4}=2.25 \\\\ &=225 \times 10^{-2} \end{aligned} $
Nearest Integers $=225$
On complete combustion 0.30 g of an organic compound gave 0.20 g of carbon dioxide and 0.10 g of water. The percentage of carbon in the given organic compound is _____________. (Nearest integer)
Explanation:
${C_x}{H_y} + \left( {x + {y \over 4}} \right){O_2} \to x\,C{O_2} + {y \over 2}{H_2}O$
Given organic compound CxHy= 0.3 gm
Produced carbon dioxide (CO2) = 0.2 gm
Produced water (H2O) = 0.1 gm
Moles of CO2 = ${{0.2} \over {44}}$
$\therefore$ Moles of C atom = ${{0.2} \over {44}}$
$\therefore$ Mass of C atom = ${{0.2} \over {44}}$ $\times$ 12 = 0.0545
Moles of H2O = ${{0.1} \over {18}}$
$\therefore$ Moles of H atoms = ${{0.1} \over {18}}$ $\times$ 2
$\therefore$ Mass of H atoms = ${{0.1} \times2\over {18}}$ $\times$ 1 = 0.0111
$\therefore$ % of C atom = ${{0.0545} \over {0.3}}$ $\times$ 100 = 18%
A protein 'A' contains 0.30% of glycine (molecular weight 75). The minimum molar mass of the protein 'A' is __________ $\times$ 103 g mol$-$1 [nearest integer]
Explanation:
Let, molar mass of protein A = x
Protein A contains 0.30% glycine
$\therefore$ ${{x \times 0.3} \over {100}} = 75$
$\Rightarrow$ x = 25000 = 25 $\times$ 103
The number of N atoms in 681 g of C7H5N3O6 is x $\times$ 1021. The value of x is (NA = 6.02 $\times$ 1023 mol$-$1) (Nearest Integer)
Explanation:
First, determine the molar mass of the compound C$_7$H$_5$N$_3$O$_6$:
Carbon (C): 12 g/mol, and there are 7 C atoms: $7 \times 12 = 84 \text{ g/mol}$
Hydrogen (H): 1 g/mol, and there are 5 H atoms: $5 \times 1 = 5 \text{ g/mol}$
Nitrogen (N): 14 g/mol, and there are 3 N atoms: $3 \times 14 = 42 \text{ g/mol}$
Oxygen (O): 16 g/mol, and there are 6 O atoms: $6 \times 16 = 96 \text{ g/mol}$
Calculate the total molar mass:
$ \text{Molar mass of C}_7\text{H}_5\text{N}_3\text{O}_6 = 84 + 5 + 42 + 96 = 227 \text{ g/mol} $
Calculate the number of moles of C$_7$H$_5$N$_3$O$_6$ in 681 grams:
$ \text{Moles of C}_7\text{H}_5\text{N}_3\text{O}_6 = \frac{681 \text{ g}}{227 \text{ g/mol}} = 3 \text{ moles} $
Each molecule of C$_7$H$_5$N$_3$O$_6$ contains 3 nitrogen atoms. Thus, in 3 moles, the total moles of nitrogen atoms is:
$ \text{Moles of N atoms} = 3 \times 3 = 9 \text{ moles} $
Using Avogadro's number, calculate the number of nitrogen atoms:
$ \text{Number of N atoms} = 9 \text{ moles} \times 6.02 \times 10^{23} \text{ atoms/mole} $
$ = 54.18 \times 10^{23} = 5.418 \times 10^{24} \text{ N atoms} $
Convert this number to the form $ x \times 10^{21} $:
$ 5.418 \times 10^{24} = 5418 \times 10^{21} $
Thus, the value of $ x $ is:
$ x = 5418 $
1 L aqueous solution of H2SO4 contains 0.02 m mol H2SO4. 50% of this solution is diluted with deionized water to give 1 L solution (A). In solution (A), 0.01 m mol of H2SO4 are added. Total m mols of H2SO4 in the final solution is ___________ $\times$ 103 m mols.
Explanation:
$=0.01 \mathrm{~m} \mathrm{~mol}$.
$\mathrm{n}_{\mathrm{H}_2 \mathrm{SO}_4}$ in Final solution $=0.01+0.01$
$=0.02 \,\mathrm{m\,mol}$
$=0.00002 \times 10^3 \mathrm{m\,mol}$
The answer 0
Number of grams of bromine that will completely react with 5.0 g of pent-1-ene is ___________ $\times$ 10$-$2 g. (Atomic mass of Br = 80 g/mol) [Nearest Integer]
Explanation:
Molar mass of C5H10 = 12 $\times$ 5 + 10 = 70 gm
Given mass of C5H10 = 5 gm
$\therefore$ Moles of C5H10 = ${5 \over {70}}$
From reaction,
1 mole of C5H10 reacts with 1 mole of Br2
$\therefore$ ${5 \over {70}}$ moles of C5H10 reacts with ${5 \over {70}}$ moles of Br2
$\therefore$ Reacted Br2 = ${5 \over {70}}$ $\times$ 160 gm
= 11.428 gm
= 1142.8 $\times$ 10$-$2 gm
$\simeq$ 1143 $\times$ 10$-$2 gm
A 0.166 g sample of an organic compound was digested with conc. H2SO4 and then distilled with NaOH. The ammonia gas evolved was passed through 50.0 mL of 0.5 N H2SO4. The used acid required 30.0 mL of 0.25 N NaOH for complete neutralization. The mass percentage of nitrogen in the organic compound is ____________.
Explanation:
Millimoles of $\mathrm{NH}_{3}=30 \times 0.25=7.5$
Mass $\%$ of nitrogen $=\frac{7.5}{0.166} \times 10^{-3} \times 14 \times 100 \simeq 63 \%$
The treatment of an aqueous solution of $3.74 \mathrm{~g}$ of $\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}$ with excess KI results in a brown solution along with the formation of a precipitate. Passing $\mathrm{H}_{2} \mathrm{~S}$ through this brown solution gives another precipitate $\mathbf{X}$. The amount of $\mathbf{X}$ (in $g$ ) is ___________.
[Given: Atomic mass of $\mathrm{H}=1, \mathrm{~N}=14, \mathrm{O}=16, \mathrm{~S}=32, \mathrm{~K}=39, \mathrm{Cu}=63, \mathrm{I}=127$ ]
Explanation:
Number of moles of $Cu{(N{O_3})_2} = {{3.74} \over {187}} = 0.02$
$\mathop {2Cu{{(N{O_3})}_2}}\limits_{0.02} + 4KI \to C{u_2}{I_2} \downarrow + \mathop {{I_2}}\limits_{0.01} + 4KN{O_3}$
$\mathop {{I_2}}\limits_{0.01} + KI \to \mathop {K{I_3}}\limits_{0.01\,(Brown\,solution)} $
$\mathop {K{I_3}}\limits_{0.01} + {H_2}S \to KI + \mathop S\limits_{0.01\,(X)} \downarrow + 2HI$
Number of moles of sulphur precipitated (X) = 0.01
Mass of sulphur precipitates (X) = 0.01 $\times$ 32 = 0.32 gm
| Compound | Weight % of $\mathrm{P}$ | Weight % of $\mathrm{Q}$ |
|---|---|---|
| 1 | 50 | 50 |
| 2 | 44.4 | 55.6 |
| 3 | 40 | 60 |
50 g of a substance is dissolved in 1 kg of water at $+90^{\circ} \mathrm{C}$. The temperature is reduced to $+10^{\circ} \mathrm{C}$. The density is increased from 1.1 to $1.15 \mathrm{~g} \mathrm{~cc}^{-1}$. What is the % change of molarity of the solution?
The empirical formula of calgon is
The statement related to law of definite proportions is
What are $x$ and $y$ in the following reaction?
$x \mathrm{~Pb}_3 \mathrm{O}_4 \longrightarrow y \mathrm{PbO}+\mathrm{O}_2$
A certain metal sulphide, MS2, is used extensively as a high temperature lubricant. If MS2 is 40.06% by mass sulphur, metal M has atomic mass.
[Given : NA = 6.02 $\times$ 1023 mol$-$1
Atomic mass of Na = 23.0 u]
Explanation:
Molar mass of Na = 23 gmol$-$1
${{Weight\,of\,sodium\,atom} \over {Molecular\,mass\,of\,sodium\,atom}} = {{Number\,of\,atoms} \over {Avogadro's\,number}}$
${{8g} \over {23g}} = {{Number\,of\,atoms} \over {6.022 \times {{10}^{23}}}}$
Number of atoms $ = {{8 \times 6.022} \over {23}} \times {10^{23}}$
Number of atoms = 2.09 $\times$ 1023
x $\approx$ 2
Hence, answer is 2.
[Atomic masses Cu : 63.54u, S : 32u, O : 16u, H : 1u]
Explanation:
The concentration of copper sulphate solution is x $\times$ 10$-$3 mol/L.
Molarity = ${{Number\,of\,moles\,of\,solute} \over {Volume\,of\,solution(L)}}$ ..... (i)
Molar mass of CuSO4 . 5H2O = 63.54 + 32 + 16 $\times$ 4
= 5 $\times$ 18 = 249.54 g/mol
Number of moles of solute = ${{Weight\,of\,solute} \over {Molecular\,mass\,of\,solute}}$
= ${{80g} \over {249.54g/mol}}$ = 0.32 mol
Volume of solution = 5 L
From Eq. (i),
Molarity = ${{0.3205} \over 5}$ = 64.11 $\times$ 10$-$3 mol/L
$\therefore$ x = 64.11
or x $\approx$ 64
Hence, answer is 64.
Explanation:
${{20} \over {62}}$ moles
Moles of NaOH formed = ${{20} \over {62}}$ $\times$ 2
[NaOH] = ${{{{40} \over {62}}} \over {{{500} \over {1000}}}}$ = 1.29 M = 13 $\times$ 10$-$1 M (Nearest integer)
Explanation:
$ \Rightarrow x \times {10^{ - 2}} = {{6.3/126} \over {250/1000}}$
$x = 20$
[Atomic weight : H = 1.008; C = 12.00; O = 16.00]
Explanation:
| t = 0 | 2.27 mole | 31.25 mol | ||
|---|---|---|---|---|
| t = $\infty $ | 0 | 19.9 mol | 6.81 mol | 9.08 mol |
mole fraction of CO2 in the final reaction mixture (heterogenous)
${X_{C{O_2}}} = {{6.81} \over {19.9 + 6.81 + 9.08}}$
= 0.1902 = 19.02 $\times$ 10$-$2
$\Rightarrow$ 19
[Atomic mass : Ag = 108, Br = 80]
Explanation:
$\Rightarrow$ nBr = nAgBr = 0.001 mol
$\Rightarrow$ massBr = (0.001 $\times$ 80) gm = 0.08 gm
$\Rightarrow$ mass% = ${{0.08 \times 100} \over {0.2}} = 40\% $
[Atomic Masses - Na : 23.0 u, O : 16.0 u, P : 31.0 u]
Explanation:
$ = {{{1 \over 3} \times {{3.45} \over {23}}mol} \over {0.1\,L}}$
= 0.5 = 50 $\times$ 10$-$2
Explanation:
1000 kg solvent $\to$ 3.3 $\times$ 74.5 gm KCl $\to$ 245.85
Weight of solution = 1245.85 gm
Volume of solution = ${{1245.85} \over {1.2}}$ ml
So molarity = ${{3.3 \times 1.2} \over {1245.85}} \times 1000 = 3.17$
The above reaction is carried out in a vessel starting with partial pressure PSO2 = 250 m bar, PO2 = 750 m bar and PSO3 = 0 bar. When the reaction is complete, the total pressure in the reaction vessel is _______ m bar. (Round off of the nearest integer).
Explanation:
$\therefore$ Final total pressure = 625 + 250 = 875 m bar
[Use : Atomic mass : Na : 23.0 u, O : 16.0 u, H : 1.0 u, Density of H2O : 1.0 g cm$-$3]
Explanation:
mass of solution = (1.2 $\times$ 1000)g = 1200 gm
Neglecting volume of NaOH
Mass of water = 1000 gm
$\Rightarrow$ Mass of NaOH = (1200 $-$ 1000)gm = 200 gm
$\Rightarrow$ Moles of NaOH = ${{200g} \over {50g/mol}} = 5$ mol
$\Rightarrow$ molality = ${{5mol} \over {1kg}} = 5$ m
Explanation:
Explanation:
M $\times$ 10 $\times$ 1 = 0.02 $\times$ 15 $\times$ 6
$ \Rightarrow $ M = 0.18 = 18 $\times$ 10$-$2 M
Explanation:
Moles of ${H_2}O = {{72} \over {18}} = 4$
Moles of C4H10 used $ = {1 \over 5} \times 4$
Wight of C4H10 used $ = {4 \over 5} \times 58$
= 46.4 gm = 464 $ \times $ 10-1 g
[Given : Atomic mass of C = 12, H = 1, O = 16 u]
Explanation:
Explanation:
Moles of NaOH = ${x \over {40}}$
Mass of Na2CO3 = y
Moles of Na2CO3 = ${y \over {106}}$
${x \over {40}} = {y \over {106}}$
x + y = 4
x = 1.1, y = 2.9
x = 1.1 $ \approx $ 1 (nearest integer)
(NA = 6.022 $\times$ 1023)
Explanation:
So, millimoles of NaOH = 250 $\times$ 0.5 = 125
Millimoles of HCl = 500 $\times$ 1 = 500
Now, reaction is
125 millimoles of NaOH reacts with 125 millimoles of HCl. So, millimoles of HCl left = 375
Moles of HCl = 375 $\times$ 10$-$3
Number of HCl molecules
= Avogadro's constant (NA) $\times$ moles of HCl
= 6.022 $\times$ 1023 $\times$ 375 $\times$ 10$-$3
= 225.8 $\times$ 1021 = 226 $\times$ 1021
Therefore, answer is 226.
(Use : $\Delta$CH(glucose) = $-$2700 kJ mol$-$1)
Explanation:
so mole of glucose needed for 104 kJ energy
= ${{10000} \over {2700}} = 3.703$ moles
Weight of glucose = 3.703 $\times$ 180 g/moles
= 666.666
$\approx$ 667 g
Hence, amount of glucose required is 667 g.
4.8 mL, 4.9 mL, 5.0 mL, 5.0 mL and 5.0 mL
Based on these readings, and convention of titrimetric estimation the concentration of Na2CO3 solution is ___________ mM.
(Round off to the Nearest Integer).
Explanation:
Na2CO3 + 2 HCl $ \to $ 2 NaCl + CO2 + H2O
n factor of Na2CO3 = 2
n factor of HCl = 1
equivalent of Na2CO3 = equivalent of HCl
$ \Rightarrow $ ${{10} \over {1000}} \times 2 \times M = {5 \over {1000}} \times 1 \times 0.2$
$ \Rightarrow M = {1 \over {20}}M$
= 0.05 M
= 50 $\times$ 10$-$3 M
= 50 mM
[Given : Atomic masses : C : 12.0 u, H : 1.0 u, O : 16.0 u ]
Explanation:
$ \therefore $ Mass of acrolein = $\left[ {{x \over {74}} \times 0.64} \right] \times 56 = 78$
$ \Rightarrow x = 16.1gm \simeq 16gm$
(Given : Atomic masses : C : 12.0 u, H : 1.0 u, N : 14.0 u, Br : 80.0 u]
Explanation:
ammonium bromide formed = ${{23} \over {230}}$ = 0.1
$ \therefore $ No. of moles of bromomethane consumed
= 3 $ \times $ 0.1
= 3 $ \times $ 10–1
Explanation:
Here, ${C_2}{H_6} = 3gm = {3 \over {30}} = 0.1$ mol
From 1 mol C2H6 we get 3 mol H2O
$ \therefore $ From 0.1 mol C2H6 we get = 3 $\times$ 0.1 mol H2O
$ \therefore $ Moles of H2O = 0.3
= 0.3 $\times$ 6.023 $\times$ 1023 molecules
= 1.8069 $\times$ 1023 molecules
= 18.069 $\times$ 1022 molecules
Consider the above reaction. The percentage yield of amide product is __________. (Round off to the Nearest Integer).
(Given : Atomic mass : C : 12.0 u, H : 1.0 u, N : 14.0 u, O : 16.0 u, Cl : 35.5 u)
Explanation:
Stoichiometric moles of amide = 10$-$3 mol
Actual weight of amide = 10-3 $ \times $ 273 = 0.273 g
% yield = ${{0.210} \over {0.273}} \times 100$
= 76.9%
$ \simeq $ 77%
Explanation:
Valance Factor of $C{r_2}{O_7}^{2 - }$ = 6
Valance Factor of $F{e^{2 + }}$ = 1
mili eq. of $C{r_2}{O_7}^{2 - }$ = mili eq. of Fe2+
6(0.03 $\times$ 20) = 1(M $\times$ 15)
M = 6(0.03)4/3
M = 0.24 M
M = 24 $\times$ 10$-$2 M