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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 851. |
The density of `3M` solution of `NaCl` is `1.25 g mL^(-1)`. The molality of the solution is |
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Answer» M= 3 mol `L^(-1)` Mass of NaCl in 1 L solution `=3xx58.5 = 175.5 g ` Mass of 1 L solution `=1000xx1.25 = 1250 g` (since density =1.25 g `mL^(-1)`) Mass of water in solution `=1250 - 75 .5` =1074.5 g Molality `=("No. of moles of solute ")/("Mass of solvent in kg ") ` `=( 3 mol )/(1.0745 kg )=2.79 m` Often in a chemistry laboratory, a solution of a desired concentration is prepared by diluting a solution of known higher concentration. The solution of higher concentration is also known as stock solution. Note that the molality of a solution does not change with temperature since mass remains unaffected with temperature. |
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| 852. |
A solution is prepared by adding 2 g of a substance A to 18 g of water. Calculate the mass per cent of the solute. |
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Answer» Mass per cent of A `=(" Mass of A ")/( " Mass of solution ")xx100` `= ( 2g)/( 2 g of A+ 18 g of "Water")xx100` `= (2 g)/( 20 g) xx 100` `10 % ` |
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| 853. |
What volume of 90% alcohol by weight `(d=0.8 g mL^(-1))` of ethanol must be used to prepare 40 mL of 10% alcohol by weight `(d=0.9 g mL^(-1))` |
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Answer» Correct Answer - 9 Let x of `Na_(2)CO_(3)` . Then, weight of `NaHCO_(3)=(15-x)g` Moles of `NaCl` is produced `=(11.0 g)/(58.5 g)=0.188` mol The `NaCl` is produced by the reaction of `(x/106)` mol of `Na_(2)CO_(3)` and `((15-x))/84` mol of `NaHCO_(3)`. Each mol of `Na_(2)CO_(3)` produces 2 mol of `NaCl` `:. (2x)/106+(15-x)/84=0.188` Solve `x := 1.35 g Na_(2)CO_(3)`, `NaHCO_(3)=(15-1.35)=13.6 g` `% Na_(2)CO_(3)=13.5/15xx100=9.0% Na_(2)CO_(3)`. |
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| 854. |
Two students performed the same experiment separately and each one of them recovered two readings of mass which are given below. Correct reading of mass is 3.0 g. On the basis of given data, mark the correct optioin out of the following statements. A. Results of both the students are neither ac curate nor preciseB. Results of student A are both percies and accurateC. Results of student B are neither precise nor accurateD. Results of student B are both precise and accurate. |
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Answer» Average of readings of student, `A=(3.01+2.99)/(2)=300` Average of readings of student ,B`=(3.05+2.95 )/(2)=300` Correct reading = 3.00 For both studnets average value is close to the correct value. Hence readings of both are accurate Readings of student A are close to each other (differ only by 0.02) and also close to the correct reading, of A are precise also. But readings of B are not close to each other (differ by 0.1) and hence are not precise. |
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| 855. |
`1 L` of `0.1 M NaOH, 1 L` of `0.2 M KOH`, and `2 L` of `0.05 M Ba (OH)_(2)` are mixed togther. What is the final concentration of the solution.A. 0.01 MB. 0.01 NC. 0.1 ND. 0.001 M |
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Answer» Total volume `(V_4)=4L` n factor for NaOH and KOH is 1 whereas that for `Ba(OH)_2` is 2 Now, `N_1V_1+N_2V_2+N_3V_3=N_4V_4` `0.1xx1+0.2xx1+0.05xx2xx2=N_4xx4` `rArr 0.1 + 0.2 + 0.2 = N_4xx4 rArr N_4 = 0.125` Hence, final concentration of solution = 0.1 N |
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| 856. |
Two students performed the same experiment separately and each one of them recovered two readings of mass which are given below. Correct reading of mass is 3.0 g. On the basis of given data, mark the correct optioin out of the following statements. A. Results of both the students are neither accurate nor precise.B. Results of student A are both precise and accurate.C. Results of student B are neither precise nor accurate.D. Results of student B are both precise and accurate. |
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Answer» Sudent A : Average reading `=(3.01+2.99)/(2)=3.0g` Student B: Average reading `=(3.05+2.95)/(2)=3.0g` For both the students A and B, average reading is close to the correct reading (i.e., 3.0g). Hence, both recorded accurate reading. But the reading recorded by student A are more precise as they differ only by`pm0.01`, whereas reading recorded by the student B are differ by `pm 0.05`. Thus, the result of student A are both precise and accurate. |
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| 857. |
What is the concentration of sugar `(C_(12)H_(22)O_(11))` in `mol L^(-1)` if its `20 g` are dissolved in enough water to make a final volume up to `2L`? |
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Answer» molarity (M) of a solution is given by , `=("Number of moles of solute")/("Volume of solution in Litres") ` `=("Mass of suger / molar mass of sugar")/(2L)` `=(20g//[(12xx12)+(1xx22)+(11xx16)]g)/(2L)` `=(20 g//342g)/(2L)` `=(0.0585 mol)/(2L)` =0.2925 mol `L^(-1)` `therefore ` Molar concentration of sugar `=0.02925 mol L^(-1)` |
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| 858. |
A compound contains elements X and Y in 1 : 4 mass ratio. If the atomic masses of X and Y are in ratio 1 : 2, then empirical formula will beA. XYB. `XY_(2)`C. `X_(2)Y`D. `X_(4)Y` |
| Answer» Correct Answer - B | |
| 859. |
Metal chloride contains 71% chlorine. Then calculate equivanlent weight of that metal bromide (at. Wt. of Br = 80)A. `14.5`B. 85C. `94.5`D. 100 |
| Answer» Correct Answer - C | |
| 860. |
A `100 ml` solution of `0.1 N HCl` was titrated with `0.2`? `N NaOH` solution. The titration. The remaining titration war completed by adding `0.25 N KOH` solution. The volume of `KOH` required for completing the titration isA. (a)` 70 ml`B. (b)` 32 ml`C. (c )` 35 ml`D. (d)` 16 ml` |
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Answer» Correct Answer - D Volume m of HCl neutralised by `NaOH` `=("Caustic soda")=V_(1)` `N_(1)V_(1)=N_(2)V_(2), 0.1xxV_(1)=0.2xx60, V_(1)=60 ml` V total (HCl)=100 ml `V_(1)=60 ml` `=40 ml` `40 ml 0.1 N HCl` is now netralised by `KOH (0.25N) rarr (HCl)` `N_(1)V_(1)=N_(2)V_(2) (KOH)` `0.1xx40=0.25xxV_(2), V_(2)=16 ml`. |
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| 861. |
Which of the following has there significant figures?A. `6.60xx10^(-30)`B. `1.70`C. `0.28`D. `6.02xx10^(23)` |
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Answer» Correct Answer - A::B::D Factual question |
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| 862. |
What volume of hydrogen gas at 273 K and 1 atm. Pressure will be consumed in obtaining 21.6 g elemental boron (Atomic mass=10.8) from the reduction of boron trichloride by hrogen?A. `89.6 L`B. `67.2L`C. `44.8L`D. `22.4L` |
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Answer» Correct Answer - B `2BCl_(3)+underset("at N.T.P.")underset(3xx22.4L)(3H_(2))tounderset(2xx10.8g)(2B)+6HCl` To produced 21.6 g `(10.8xx2g)` of boron hydrogen required at S.T.P. `= 3xx22.4L = 67.2L` |
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| 863. |
In the reaction: `Na_(2)S_(2)O_(3)+4Cl_(2)+5H_(2)O rarr Na_(2)SO_(4)+H_(2)SO_(4)+8HCl`, the equivalent weight of `Na_(2)S_(2)O_(3)` will be: (M= molecular weight of `Na_(2)S_(2)O_(3))`A. (a)`M//4`B. (b)`M//8`C. (c )`M//1`D. (d)`M//2` |
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Answer» Correct Answer - B `Na_(2) overset(+2)(S_(2) O_(3)) rarr Na_(2) overset(+6)(S O_(4))` the total change in oxidation number `=4xx2=8` `:. E_(Na_(2)S_(2)O_(3))=(mol. Wt.)/(v.f)=M/8` |
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| 864. |
`28 NO_(3)^(-)+3As_(2)S_(3)+4H_(2)O rarr 6AsO_(4)^(3-)+28NO+9SO_(4)^(2-)+H^(+)` What will be the equivalent mass of As_(2)S_(3)` in the above reaction?A. (a)`M/2`B. (b)`M/4`C. (c )`M/24`D. (d)`M/28` |
| Answer» Correct Answer - D | |
| 865. |
For the redox reaction, `MnO_(4)^(-) + C_(2)O_(4)^(2-) + H^(+) rarr Mn^(2+) + CO_(2) + H_(2)O` the correct coefficients of the reactants for the balanced reaction areA. `{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),(16,5,2):}`B. `{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),(2,5,16):}`C. `{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),(2,16,5):}`D. `{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),(5,16,2):}` |
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Answer» Correct Answer - B `overset((+7))(Mn)O_(4)^(-) rarr Mn^(+2), 5e^(-)` gain `overset((+3))(C_(2))O_(4)^(-2) rarr overset((+4))(CO_(2)), 2e^(-)` loss Multiplying (1) by 2 and (2) by 5 to balance `e^(-)` `2MnO_(4)^(-)+5 C_(2)O_(4)^(-2) rarr 2 Mn^(+2)+10 CO_(2)` on balancing charge , `2MnO_(4)^(-)+5C_(2)O_(4)^(-2)+16H^(+) rarr 2Mn^(+2)+10 CO_(2)+8 H_(2)O` |
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| 866. |
Which of the following statement regarding Avogadro number is(are) correct?A. It is `6.023 xx 10^(23)`B. It is the number of atoms present in exactly `12g` of `C-12` isotopeC. It is the number of atoms present in `1.0` mole on any substanceD. It is the number of atoms of deuterium present in its `2.0g` |
| Answer» Correct Answer - A::B::D | |
| 867. |
The number of atoms present in one mole of an element is equal to Avogadro number. Which of the following element contains the greatest number of atom?A. 4g HeB. 46g NaC. 0.40g CaD. 12 g He |
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Answer» For comparing number of atoms, first we calcualte the moles as all are monoatomic and hence, moles `xxN_(A)`= number of atoms . Moles of `4g He=(4)/(4)=1` mol `46g Na=(46)/(23)=2` mol `0.40g Ca=(0.40)/(20)=0.1` mol `12g He=(12)/(4)=3` mol Hence 12 g He contains greatest number of atoms as it possesses maximum number of moles. |
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| 868. |
If the concentration of glucose `(C_(6)H_(12)O_(6))` in blood is 0.9 g `L^(-1)`, what will be the molarity of glucose in blood?A. 5 MB. 50 MC. 0.005 MD. 0.5 M |
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Answer» Molarity `=("W_B (in g)")/(M_(B)("in g mol"^(-1))xx "Volume of solution (in L)")` `=(Conc. ("in g L"^(-1)))/(M_(B) ("in g mol"^(-1)))= (0.9)/(180) = 0.005M (therefore " molar mass of " C_6H_12O_6= 180 g mol^(-1))` |
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| 869. |
Assertion: `31.26 mL` of `0.165 M` solution of `Ba(OH)_(2)` is exactly neutralised by `25 mL` of citric acid `C_(6)H_(8)O_(7)` of molarity `0.138`. Reason: The acid is tribasic in nature.A. (a)If both assertion and reason are true and the reason is the correct explanation of the assertion.B. (b)If both assertion and reason are true and the reason is not the correct explanation of the assertion.C. (c )If assertion is true but reason is false.D. (d)If assertion is false but reason is true. |
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Answer» Correct Answer - A Meq. Of `Ba(OH)_(2)`=Meq. Of acid `31.26xx0.165xx2=25xxMxxn=25xx0.138xxn` `:. N=3` |
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| 870. |
If the concentration of glucose `(C_(6)H_(12)O_(6))` in blood is 0.9 g `L^(-1)`, what will be the molarity of glucose in blood?A. 5MB. 50MC. 0.005MD. 0.5M |
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Answer» In the given question 0.9 `gL^(-1)` means that 1000(mL) solution contains 0.9g of glucose. `therefore` Number of moles `=0.9g "glucose"=(0.9)/(180)` mol glucose `=5xx10^(-3)` mol glucose (where , molecular mass of glucose `(C_(6)H_(12)O_(6))=12xx6+12xx1+6xx16=180u`) i.e. L solution contains 0.05 mole glucose or the molarity of glucose is 0.005M. |
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| 871. |
In a test-tube, there is 18 g of glucose `(C_(6)H_(12)O_(6))` 0.08 mole of glucose is taken out. Glucose left in the test tube isA. 0.10 gB. 0.02 gC. 0.10 molD. 3.60 g |
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Answer» Correct Answer - D Molar mass of glucose, `C_(6)H_(12)O_(6) = 180 g mol^(-1)` 18 g glucose = 0.10 mol Moles of glucose left = 0.10-0.08 = 0.02 mol Mass of glucose = `(0.02 mol)xx(180 g mol^(-1)) = 360g` |
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| 872. |
Which of the following data illustates the law of conservation of mass?A. 56 g of CO reacts with 32 g oxygen to produce 44 g of `CO_(2)`B. 1.70 g of `AgNO_(3)` reacts with 100 mL of 0.1 MHCl to produce 1.435 g of AgCl and 0.63 g of `HNO_(3)`.C. 12 g of C is heated in vaccume and on cooling there is no change on mass.D. None of the above. |
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Answer» Correct Answer - B Total mass of reactants `=1*70+(100xx0*1xx10^(-3)xx36*5)` `=2*065 g` Mass of products `=0*63+1*435=2*065g` Hence, law of conservation of mass is proved. |
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| 873. |
If law of conservation of mass was to hold true, then `20*8 g` of `BaCl_(2)` on reaction with `9*8 g` of `H_(2)SO_(4)` will produce `7*3` of HCl and `BaSO_(4)` equal toA. `11*65 g`B. `23*3 g`C. `25*5 g`D. `30*6 g` |
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Answer» Correct Answer - B `underset(208g)(BaCl_(2))+ underset(98g)(H_(2)SO_(4))tounderset(233 g)(BaSO_(2))+underset(73 g)(2HCl)` |
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| 874. |
How many moles of lead (II) choride will be formed from a reaction between 6.5 g of PbO and 3.2 g of HCl?A. (a)`0.333`B. (b)`0.011`C. (c )`0.029`D. (d)`0.044` |
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Answer» Correct Answer - c `PbO+2HCl rarr PbCl_(2) rarr PbCl_(2)+H_(2)O` `n_(PbO)=6.5/223=0.029` `n_(HCl)=3.2/36.5=0.087` `{:(PbO " gives PbCl"_(2),|,"HCL gives" ,PbCl_(2),),("Moles " 0.029,0.087,0.048,,),(,impliesn_(PbCl_(2)),=0.029,,):}` |
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| 875. |
What will be the molarity of the solution in which 0.365 g of HCl gas is dissolved in 100 mL of solution ?A. 2 MB. 0.2 MC. 1 MD. 0.1 M |
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Answer» No. of moles in 0.365g of HCl `=(0.365)/(36.5)=0.01` Volume of solution in L `=100/1000=0.1L` Molarity `=n/(V("in L"))=(0.01)/(0.1)=0.1M` |
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| 876. |
Molarity of liquid HCl with density equal to 1.17g/cc isA. `36.5`B. `18.25`C. `32.05`D. `4.65` |
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Answer» Correct Answer - C Liquid HCl is 100 % solution of mass Thus, `M = (100xx1.17xx10)/(36.5) = 32.05M` |
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| 877. |
How many moles of HCl are present in 1 litre of 1 M HCl solution ? |
| Answer» Correct Answer - 1 mol | |
| 878. |
Calculate the molality of a solution containing 8 moles of sucrose in 250 g of solvent. |
| Answer» Correct Answer - 32 mol `kg^(-1)` | |
| 879. |
A solution contains 10 moles of sucrose in 1 kg of solvent. Calculate the molality of solution. |
| Answer» Molality (m) `=("No. of moles of solute")/("kg of solvent")=(10 mol)/(1 kg)=10 mol kg^(-1)` | |
| 880. |
Assertion: Molality and mole fraction concentration units do not change with temperature. Reason: These units are not defined in terms of any volume.A. (a)If both assertion and reason are true and the reason is the correct explanation of the assertion.B. (b)If both assertion and reason are true and the reason is not the correct explanation of the assertion.C. (c )If assertion is true but reason is false.D. (d)If assertion is false but reason is true. |
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Answer» Correct Answer - A Due to temperature change volume get changed. Hence concentration units dependent on volume get changed. |
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| 881. |
Density of `2.05 M` solution of acetic acid in water is `1.02g//mL`. The molality of same solution is:A. (a)0.44 mol `kg^(-1)`B. (b)1.14 mol `k^(-1)`C. (c )3.28 mol `kg^(-1)`D. (d)2.28 mol `kg^(-1)` |
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Answer» Correct Answer - D weight of acetic acid=`2.05xx60=123 g` Weight of solution`=1000xx1.02=1020 g` `:.` Weight of water`=(1020-123)=897 g` `:.` Molality`=(2.05xx1000)/897=2.285` |
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| 882. |
Calculate the molality of aq. Glucose solution in which mole fraction of glucose is 0.05. |
| Answer» Correct Answer - `2.92` | |
| 883. |
A solution is prepared by adding 360 g of glucose to 864 g of water. Calculate mole fraction of glucose (molar mass of glucose = 180). |
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Answer» No. of moles of glucose `=(360)/(180)=2` No. of moles of water `=(864)/(18)=48` `therefore` Mole fraction of glucose `=(2)/(2+48)=(2)/(50)=(1)/(25)=0.04` |
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| 884. |
Calculate the molality of solution containing 3 g glucose dissolved in 30 g of water . (molar mass of glucose = 180)A. 0.50 mB. 0.56 mC. 0.091 mD. 0.05 m |
| Answer» Correct Answer - B | |
| 885. |
A solution is prepared by adding 64 g of `CH_(3)OH` to 180 g of water. Calculate the mole fraction of `CH_(3)OH`. (Molar mass of `CH_(3)OH = 32 g`) . |
| Answer» Correct Answer - `0.166` | |
| 886. |
Molecular mass of `C_(3) H_(7) OH` isA. 60 uB. 40 uC. 30 uD. 50 u |
| Answer» Correct Answer - A | |
| 887. |
Calculate percentage of carbon in ethanol `(C_(2)H_(5)OH)`. |
| Answer» Correct Answer - `52.174%` | |
| 888. |
What is the mass percent of oxygen in ethanol ?A. 0.5214B. 0.1313C. 0.16D. 0.3473 |
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Answer» Molecular formula of ethenol `=C_2H_5OH` Molar mass of ethenol of `2xx12.01+6xx1.008+16` =46.068g Mass percent of oxygen `=(16)/(46.068)xx100=34.73%` |
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| 889. |
How much mass of sodium acetate is required to make 250 mL of 0.575 molar aqueous solution?A. 11.79 gB. 15.38 gC. 10.81 gD. 25.35 g |
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Answer» Molar mass of sodium acetate `(CH_3COONa)` =82.0245 g/mol Mass of `CH_3COONa` required to make 250 mL of 0.575M solution `=(0.575xx82.0245xx250)/(1000)=11.79g` |
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| 890. |
Calculate the mass of sodium acetate `(CH_(3)COONa)` required to make `500 mL` of `0.375` molar aqueous solution. Molar mass of sodium of acetate is `82.0245 g mol^(-1)`. |
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Answer» 0.375 M aqueous solution od sodium acetate `-=` 1000 mL of solution containing 0.375 moles of sodium acetate `therefore ` Number of moles of sodium aceate in 500 ml `=(0.375 )/(1000)xx500` `=0.1875 ` mole Molar mass of sodium acetate = 82.0245 g `"mole"^(-1)` (Given ) `therefore ` Required mass of sodium acetate `=(82.0245 g mol ^(-1) ) (0.1875 "mole")` =15.38 g |
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| 891. |
Calculate the amount of carbon dioxide that could be produced when a. `1` mol of carbon is burnt in air b. `1` moles of carbon is burnt in `16 g` of dioxygen. c. `2` moles of carbon are burnt in `16 g` of dioxygen. |
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Answer» the balanced reaction of combustion of carbon can be written as : (i) As per the balanced equation , 1 mole of carbon burns in mole of dioxygen (air) to produce 1 mole of carbon dioxide . (ii) According to the question ,only 16 g of dioxygen is available ,Hence , it will react with 0.5 mole of carbon to give 22 g of carbon dioxide ,Hence , it is a limiting reactant . (iii) According to the question ,only 16 g of dioxygen is available it is a limiting of carbon dioxide . |
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| 892. |
Calculate the moles of `CO_(2)` obtained when `0.274` mole Of `C_(2) H_(5)OH` is burnt in air.A. `0.548`B. `0.0548`C. `0.558`D. `0.058` |
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Answer» Correct Answer - A `underset(underset(0.274 "mol")(1"Mol"))(C_(2)H_(5)OH)+3O_(2)rarr underset(2"mol")(3 "CO")_(2)+3H_(2)O` 1 mole of `C_(2)H_(5)OH` prodluces 2 moles of `CO_(2)` `therefore 0.274` mole of `C_(2)H_(5)OH` will produce `(CO_(2)` `=2/1xx 0.274` `=2xx 0.274=0.548` mol Thus, `0.548` mole of `CO-(2)` is obtained from `0.273` mole ` C_(2) H_(5)OH.` |
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| 893. |
The molecular mass of glucose `(C_(6) H_(12) O_(6))` molecule to three significant figures is |
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Answer» Molecular mass of glucose `(C_(6)H_(12)O_(6))` `=6(12.011 u) +12(1.008 u)+ 6(16.00 u)` `=(72.066 u)+(12.096 u) + (96.00 u) ` `=180.162 u` |
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| 894. |
A compound contains `4.07% H,. 24.27% C`, and `71.65% Cl`. If its molar mass is `98.96`, the molecular formula will beA. CHClB. `CH_(3)Cl`C. `C_(2)H_(4)Cl_(2)`D. `C_(2)HCl` |
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Answer» Correct Answer - C Let us assume 100 g of compound contains `4.07` g H, `24.27` g C and `71. 65` g Cl. Number of moles of `"H" = (4.07)/1=4.07` Number of moles of `"C" = (24.27)/12=2.02` Number of moles of `"Cl"=(71.65)/35.5=2.02` Ratio of H : C : Cl : is 4 : 2 : 2 or 2 : 1 : 1 `because` Its empirical formula is `CH_(2)Cl`. Now, we have `n=("Molecular formula weight")/("Empirical formule weight") ` `=(98.96)/49.5=2` `therefore` Molecular formula `=nxx ("empirical formula")` `=2xx(CH_(2)Cl)` `=C_(2)H_(4)Cl_(2)` |
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| 895. |
A compound contains `4.07% H,. 24.27% C`, and `71.65% Cl`. If its molar mass is `98.96`, the molecular formula will be |
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Answer» Step 1. Conversion of mass per cent to grams Since we are having mass per cent, it is convenient to use 100 g of the compound as the starting material. Thus, in the 100 g sample of the above compound, 4.07g hydrogen, 24.27g carbon and 71.65g chlorine are present. Step 2. Convert into number moles of each element Divide the masses obtained above by respective atomic masses of various elements. This gives the number of moles of constituent elements in the compound Moles of hydrogen `=(4.07 g) /( 1.008 g) =4.04` Moles of carbon `=(24.27g) /(12.01g)=2.021` Moles of clorine` =(71.65 g) /(35.453 g) =2.021` Step 3. Divide each of the mole values obtained above by the smallest number amongst them Since 2.021 is smallest value, division by it gives a ratio of 2:1:1 for H:C:Cl . In case the ratios are not whole numbers, then they may be converted into whole number by multiplying by the suitable coefficient. Step 4. Write down the empirical formula by mentioning the numbers after writing the symbols of respective elements `CH_(2)Cl` is, thus, the empirical formula of the above compound. Step 5. Writing molecular formula (a) Determine empirical formula mass by adding the atomic masses of various atoms present in the empirical formula. For `CH_(2) Cl` empirical formula mass is `12.01+(2xx1.008)+35.453` `=49.48 g` (b) Divide Molar mass by empirical formula mass `("molaar mass ")/("Empirical formula mass ") =(98.96 g) /(49.48 g) ` (c) Multiply empirical formula by n obtained above to get the molecular formula Empirical formula `=CH_(2) Cl,n=2 ` Hence molecular fromula is `C_(2) H_(4) Cl_(2)` |
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| 896. |
Assertion : Gases combine in simple ratio of their volume but, not always. Reason : Gases deviate from ideal behaviour.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - A Both assertion and reason are true and reason is the correct explanation of assertion. |
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| 897. |
The equivalent mass of chlorine is 35.5, and the molar mass of copper is 63.5. The equivalent mass of copper chloride is 99.0. Hence, formula of copper chloride isA. `CuCl`B. `Cu_(2)Cl`C. `CuCl_(2)`D. `Cu_(2)Cl_(2)` |
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Answer» Correct Answer - A Equivalent mass lof `Cl = 35.5` g Equivalent mass of copper chloride ` = 99.0 ` g Equivalent of copper combined `=99. 0 - 35. 5 = 63. 5` Molar mass of copper `=63.5` Since, equivalent mass = molar mass Hence, the formula of compound `= CuCl` |
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| 898. |
What is the mass per cent of carbon in carbon dioxide?A. 0.00034B. 0.2727C. 0.034D. 0.287 |
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Answer» Molecular mass of `CO_(@)=1xx12+2xx16=44g` 1 g molecule of `CO_(2)` contains 1g atoms of carbon `because 44g "og" CO_(2)` contain C=12 g atoms of carbon `therefore % "of" C "in" CO_(2)=(12)/(44)xx100=27.27%` Hence, the mass per cent of carbon in `CO_(2)`is 27.27% |
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| 899. |
The empirical formula and molecular mass of a compound are `CH_(2)O` and 180 g respectively. What will be the molecular formula of the compound ?A. `C_(9)H_(18) O_(9)`B. `CH_(2)O`C. `C_(6)H_(12)O_(6)`D. `C_(2)H_(4)O_(2)` |
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Answer» Correct Answer - C Empirical formula mass `=CH_(2)O` `=12 + 2 xx 1 + 16 = 30` Molecular mass = 180 `n= "molecular mass"/"Empirical formula mass"` `= 180/ 30 = 6` `therefore` Molecular formula `= n xx` empirical formula ` = 6 xx CH_(2) O = C_(6) H_(12) O_(6)` |
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| 900. |
Empirical formula of a compound is `CH_(2)O` and its molecular mass is 90. The molecular formula of the compound isA. `C_(2)H_(4)O_(2)`B. `C_(3)H_(6)O_(3)`C. `C_(4)H_(8)O_(4)`D. `CH_(2)O` |
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Answer» Correct Answer - C E.F. mass `= 12+2xx1+16=30` Mol. Mass = 120 `n = ("Mol. Mass")/("E.F. mass")=(120)/(3)=4` `:.` Mol formula `= (CH_(2)O)_(4) = C_(4)H_(8)O_(4)` |
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