94420 + Interview Questions in Chemistry in Class 12 Page 1839 InterviewSolution

91901.	A 0.01m aqueous solution of AlCL_3 freezes at -0.068^@C Calculate the percentage of dissociation.
Answer» SOLUTION :USE the relation : `Delta T_f=iK_fm`, where I is van.t Hoff factor. Given that `m=0.01 , Delta T_f=0.068^@C, K_f=1.86K kg mol^-1` Substituting the values in the above equation, we have `0.068 k=1 times 86 k kg mol^-1 times 0.01 mol kg^-1` or `i=0.068/(1.86 times 0.01)3.6559`....(i) `AlCl_3 to AL^(3+)+3Cl^(-)` 1-x x 3X `i=(1+3x)/1` From (i) and (ii), we have `1+3x=3.6559 or 3x=2.6559 or x=0.8853` Percentage of dissociation =`0.8853 times 100=88.53`

Discussion

91902.	._(6)^(13)C and ._(7)^(14)N are the
Answer» Isotopes Isotones Isobars Isosteres Solution :They have IDENTICAL no. of NEUTRONS.

Discussion

91903.	A 0.010 M solution of maleic acid, amonoprotic organic acid, is 14% ionized. What is K_(a) for maleic acid
Answer» `2.3 xx 10^(-3)` `2.3 xx 10^(-4)` `2.0 xx 10^(-4)` `2.0 xx 10^(-6)` Answer :C

Discussion

91904.	6.0xx10^(20) molecules of urea are present in 100 L of his solution. The concentration of urea solution is
Answer» 0.001 M 0.1 M 0.02 M 0.01 M Solution :No. of moles of urea `= (6.02xx10^(20))/(6.02xx10^(23))=10^(-3)` MOL `:. 10^(-3)` mol of urea are present in 100 ML `:.` Moles of urea present in 1L (1000 mL) `=10^(-3)xx10=10^(-2)` `:.` Molarity of solution `= 10^(-2) M = 0.01 M`

Discussion

91905.	A 0.01 M solution of glucose in water freezes at- 0.0186^(@)C. A 0.01 M solution of NaCl in water will freeze at
Answer» `0^(@)C` `0.0186^(@)C` `-0.0186^(@)C` `-0.0372^(@)C` Solution :Depression in freezing POINT will be ALMOST DOUBLE because the NUMBER of particles of 0.01 M NaCl are double than 0.01 M glucose.

Discussion

91906.	60g ofcompound on analysis gave C= 24g, H= 4 g and O = 32g. If Empirical formula is
Answer» `C_(2)H_(4)O_(2)` `C_(2)H_(2)O` `CH_(2)O_(2)` `CH_(2)O` Solution :`{:("Element"," NO. of moles"," Simple RATIO"),(C =24,24//12=2,1),(H=4,4//1=4,2),(O=32,32//16,1):}` Therefore `CH_(2)O`.

Discussion

91907.	A 0.01 M ammonia solution is 5% ionized. The concentration of OH^- ion is:
Answer» 0.005 M 0.0001 M 0.0005 M 0.05 M Answer :C

Discussion

91908.	A 0.01 M ammonia solution is 5% ionized, the concentration of OH^(-) ion is :
Answer» 0.005 M 0.0001 M 0.0005 M 0.05 M SOLUTION :`[OH^(-)]=C alpha, c=0.01 M, alpha=(5)/(100)=0.05` `[OH^(-)]=0.01xx0.05 = 0.0005 M`

Discussion

91909.	6.02 xx 10^(20) molecules of urea are present in 100 mL of its solution. The concentration of urea solution is (Avogadro constant, N_(A) = 6.02 xx 10^(23)mol^(-1))
Answer» 0.02 M 0.01 M 0.001 M 0.1 M Glucose. Solution :`"No. of MOLES of urea"=(6.02xx10^(20))/(6.02xx10^(23))` =0.001 mol. 100 mL of solution has moles = 0.001 1000 mL of solution has moles `=(0.001)/((10ML))XX(1000mL)=0.01 M.`

Discussion

91910.	(6.023xx10^(23)xx1)/10^(20)=6023 sec=1.673 hrs. A 100% pure sample of a divalent metal carbonate weighing 2g on complete thermal decomposition releases 448 cc of carbon dioxide at STP. The equivalent mass of the metal is
Answer» 40 20 28 12 Solution :Divalent metal carbonate means `MCO_(3)`. `{:(ACO_(3) overset(Delta )(rarr)MO+CO_(2)),("1 mol22400 CC"),("2 g448 cc"),("448 cc evolves from 2g"):}` 22400 cc will evolve from `2/448xx22400g=100g` i.e., 100 G is the MOLECULAR weight of the carbonate. Then M.W. of metal `=100-(12+3xx16)` (wt. of carbonate, `CO_(3)^(2-)`) `=40 g` Equivalent weight `=("Molecular weight")/("Valency")` `=40/2=20`.

Discussion

91911.	A 0.004 M Solution of Na_(2)SO_(4) is isotonic with a 0.010 M solution of glucose at the temperature. The apparent degree of dissociation of Na_(2)SO_(4) is
Answer» `25%` `50%` `75%` `85%` Solution :`pi(Na_(2)SO_(4))=iCRT=i(0.004)RT` `pi"(Glucose) = CRT = 0.010 RT"` As solution are isotonic, I (0.004) RT = 0.01 RT. This givesI = 2.5 `{:("Now",Na_(2)SO_(4),hArr,2Na^(+),+,SO_(4)^(2-)),(,"1 MOLE",,0,,0),(,1-alpha,,2ALPHA,,alpha","):}` `"Total "=1+2alpha` `therefore""i=1+2alpha` `"or"alpha=(i-1)/(2)=(2.5-1)/(2)=0.75=75%`.

Discussion

91912.	A 0.01 D^(**) solution of KCl has a specific conductance value of 0.00141 mho cm^(-1). A cell filled with this solution has a resistance of 42156 ohms. (a) What is the cell constant ? (b) The same cell filled with a solution of Hcl has a resistance of 1.0326 ohms. What is specific conductivity of the HCl solution ?
Answer» Solution :(a) Cell constant `= ("specific conductance")/("conductance")` `= ("specific conductance")/("1/resistance")` `= (0.00141)/(1//4.2156) = 0.00594 CM^(-1)`. (b) Conductance of HCL solution `= (1)/(1.0326)` mho. Sicne the same cell is USED, we SHALL take the same value of cell constant in this case. SP. conductance = cell constant `xx` conductance `= 0.00594 xx (1)/(1.0326)` `= 0.00575 "mho cm"^(-1)`.

Discussion

91913.	6.02xx10^(20) molecules of urea are present in 100 mL of its solution. The concentration of the solution is
Answer» 0.02 M 0.01 M 0.001 M 0.1 M Solution :No. of moles of urea present in 100 mL of solution `=(6.02xx10^(20))/(6.20xx10^(23))=10^(-3)"MOL"` `therefore` Molar CONCENTRATION of urea in the solution `=(10^(-3))/(100)xx1000=10^(-2)M=0.01M`

Discussion

91914.	6.022 xx 10^(22) molecules of N_2 at NTP will occupy a volume of
Answer» 2.234 litre 6.02 ml 6.02 litre 22.4 litre Answer :A

Discussion

91915.	A 0.004 M solution of Na_(2)SO_(4) is isotomic with a 0.01 M solution of glucose at same temperature. The apparent degree of dissociation of Na_(2)SO_(4) is:
Answer» `25%` `50%` `75%` `85%` Solution :Since the SOLUTIONS are ISOTONIC, They have the same osmotic pressure also. `pi_(Na_(2)SO_(4))=pi_("glucose")` `ixx0.004 RT = 0.01 RT` `i=0.01/0.004=2.5` `Na_(2)SO_(4)hArr2Na^(+)+SO_(4)^(2-)` `ALPHA=(i-1)/(n-1)=(2.5-1)/(3-1)=1.5/2` =0.75 or 75%

Discussion

91916.	6.02 xx10^(20) molecules of urea are present in I 00 ml of its solution. The concentration of urea solution is
Answer» `0.02 M ` `0.01M` `0.001 M` `0.1 M` SOLUTION :`M =(6.02 xx10^(20) xx1000)/(6.02 xx10^(23) XX100)= 0.01M`

Discussion

91917.	6.02 xx 10^20 molecules of urea are present in 100 mL. of its solution. The concentration of solution is
Answer» 0.1M 0.001M 0.01M 0.02M Answer :C

Discussion

91918.	A 0.004 M solution of Na_(2)SO_(4) is isotonic with 0.010 M solution of glucose at same temperature. The percentage dissociation of Na_(2)SO_(4) is
Answer» 0.25 0.5 0.75 0.85 Solution :`{:(Na_(2)SO_(4) ,hArr, 2Na^(+)+, SO_(4)^(2-)),((0.004-X),, 2X, x):}` Since both SOLUTIONS are ISOTONIC, therefore, `0.004 + 2x = 0.01, x = 3 xx 10^(-3)` `:.` % DISSOCIATION `= (3 xx 10^(-3))/(0.004) xx 100 = 75`

Discussion

91919.	6.02 xx 10^20molecules of urea are present in 100ml of its solution. The concentration of solution is
Answer» 0.02 M 0.01 M 0.001 M 0.1 M SOLUTION :MOLES of urea ` = (6.02 xx 10^20)/(6.02 xx 10^23) = 10^(-3)` moles Molarity ` = (10^(-3) )/(100) xx 1000 = 0.01 M`

Discussion

91920.	600 mL of aqueoussolutioncontining2.5g of a protein shows an osmotic pressureof 25mm Hg at 27^(@)C.Determinethe molecular mass of protein.
Answer» Solution :`pi = cRT =(w_(2))/(M_(2)V) RT or M_(2) = (w_(2)RT)/(piV)` `w_(2) = 2.5 g , R = 0.0821 L atm mol^(-1) K^(-1)` ` T= 273 + 27 = 300 K` ` pi = (25)/(760) atm ,V = (600)/(1000) L` `M_(2)= ((2.5 g xx 0.0821 L atm mol^(-1)K^(-1)) xx (300K))/(((25)/(760) atm) xx((600)/(1000)L))` ` = 3119.8 g mol^(-1)`

Discussion

91921.	A 0.004M solution of Na_(2)SO_(4) is isotonic with a 0.010 M solution of glucose at same temperature. The apparent degree of dissociation of Na_(2)SO_(4) is
Answer» `25%` `50%` `75%` `85%` ANSWER :C

Discussion

91922.	60 mL of a mixture of nitrous oxide and nitric oxide was exploded with excess of hydrogen. If 38 mL of N_(2) was formed calculated the volume of each gas in the mixture
Answer» SOLUTION :`V_(NO)=44 mL` `V_(N_(2)O)=16 mL`

Discussion

91923.	6.0 g of urea (molecular weight = 60) was dissolved in 9.9 moles of water. If the vapour pressure of pure water is p_(0), the vapour pressure of solution is
Answer» `0.10 p_(0)` `1.10 p_(0)` `0.90 p_(0)` `0.99 p_(0)` Solution :Mole of urea `= (6)/(60) = 0.10` ACCORDING to Raoult's law `(p_("solvent") - p_("solution"))/(p_("solvent")) = x_("SOLUTE") = (n_(A))/(n_(A) + n_(B))` Hence, `(p_(0) - p_("solution"))/(p_(0)) = (0.10)/(0.10 + 9.9)` or `p_(0) - p_("solution") = 0.01 p_(0)` or `p_("solution") = 0.99 p_(0)`

Discussion

91924.	6 moles of 'A' and 10 moles of 'B' are mixed and allowed to react according to the equation. A+3Brarr2C How many moels of C are present when there are 4 moles of A in the container?
Answer» ANSWER :4

Discussion

91925.	A 0.0020 m aqueous solution os an ionic compound [Co(NH_(3))_(5)(NO_(2))] CI freezes at -0.0732^(@)C. Number of moles of ions which 1 mole of an ionic compound produces on being dissolved in water will be:
Answer» 3 4 1 2 Solution :`DeltaT_(f)=i K_(f)m` `i=((0.0732K))/((1.86Km^(-1))xx(0.0020m))=2` Since, the ionic compound complete dissociates in aqueous OLUTION, DEGREE of dissociation `(PROP)is ~~1`. For dissociation, `ALPHA=(i-1)/(n-1)` `1=(2-1)/(n-1)or n=1+1=2` No. of mole ions formed=2

Discussion

91926.	6 moles of an ideal gas expand isothermally and reversibly from a volume of 1 litre to a volume of 10 litres at 27^(@)C. What is the maximum work done
Answer» 47 kJ 100 kJ 0 34.465 kJ Solution :WORKDONE in isothermal reversible expansion `W=PDeltaV` `W=-2.303 nRTlog.(V_(2))/(V_(1))` `=-2.303xx6xx8.314xx300log.(10)/(1)` =34464.8 JOULE = 34.465 kJ.

Discussion

91927.	6 mole of ethyl alcohol reacts with sodium metal. How many moles of hydrogen are liberated ?
Answer» 2 3 4 6 Answer :B

Discussion

91928.	A 0.0020m aqueous solution of an ionic compound Co(NH_(3))_(5)(NO_(2))Clfreezes at -0.00732^(@)C. Number of moles of ions which 1 mol of ionic comound produces on being dissolved in water be (K_(f)=-1.86^(@)C//m)
Answer» 2 3 4 1 Solution :`DELTA T_(F)= ixxk_(f)xxm` `i=(Delta T_(f))/(k_(f)xxm)=(0.00732)/(1.86xx0.002)=2`.

Discussion

91929.	A 0.0020 m aqueous solution of an ionic compound Co(NH_(3))_(5) (NO_(2))Cl freezes at -0.00744^(@)C. The number of moles of ions which 1 mole of ionic compound produces on being dissolve in water is (K_(f)=-1.86^(@)C//m)
Answer» SOLUTION :No. of moles of ions produce from 1 mole of ionic COMPOUND = i `because T_(F)=K_(f)i.m.` `therefore i=(T_(f))/(K_(f)xxm)=(0.00744)/(1.86xx0.0020)=2`.

Discussion

91930.	A 0.0020 M aqueous solution of an ionic compound Co(NH_(3))_(4)(NO_(2))_(2)Cl freeze at - 0.00732^(@)C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be (K_(f)=1.86^(@)C//m).
Answer» 3 4 1 2 Solution :The NUMBER of moles produced by 1 MOLE of ionic compound `DeltaT_(F)=iK_(f)m, i=DeltaT_(f)//K_(f)m` `DeltaT_(f)=0-(-0.00732)` `=0.00732^(@)C` `K_(f)=1.86^(@)C//m` `m=0.0020` `i=(0.00732)/(1.86xx0.0020)` `=1.96~~2`

Discussion

91931.	A 0.0020 m aqueous solution in an ionic compound Co(NH_3)_5 (NO_2)Clfreezes at -0.00732^@C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be (K_f=+1.86^@ C/m)
Answer» 1 2 3 4 Answer :B

Discussion

91932.	6 grams of a substance 'A' dissolved in 100 g of water freezes at - 0.93^(@)C. Themolecular mass of 'A' is (K_(f)=1.86 Km^(-1)).
Answer» 60 120 180 140 Solution :`DELTA T_(F)=0-(-0.93)=0.93` `M_(B)=(K_(f)xx1000xx w_(B))/(w_(A)XX Delta T_(f))` `=(1.86xx1000xx6)/(100 xx 0.93)=120`

Discussion

91933.	A 0.001 molal solution of a complex with molecular formula [Pt(NH_(3))_(4)Cl_(4)] in water showed a freezing point depression of 0.0054^(@)C. If K_(f) for water is 1.80, what is the correct formulation of the molecule with a proper coordination sphere?
Answer» SOLUTION :Suppose `[Pt(NH_(3))_(4)Cl_(4)]OVERSET("Dissociation")RARR" n moles of product ions "therefore""i=n` Substituting the given data in the formula, `DeltaT_(f)=iK_(f)m`, we get `0.0054=nxx1.80xx0.001"or"n=3` Hence, the formula MUST be one which gives 3 ions of the products, i.e., `[Pt(NH_(3))_(4)Cl_(2)]Cl_(2)` `([Pt(NH_(3))_(2)Cl_(2)]Cl_(2)rarr[Pt(NH_(3))_(4)Cl_(2)]^(2+)+2Cl^(-))`

Discussion

91934.	6 gm nitrogen on successive reaction with different compounds gets finally converted into 30 gm [CR(NH_(3))_(x)Br_(2)] value of x is [ Atomic mass of Cr = 52, Br =80]
Answer» ANSWER :4

Discussion

91935.	A 0-05 M sodium hydroxide offered a resistance of 31.6Omega in a conductivity cell at 298 K. Calculate the molar conductance (cell constant of cell =0367cm^(-1)).
Answer» Solution :Specific conductance (K) `=("CELL CONSTANT")/(R)` `R=31.6Omega` Cell constant `=0.367cm^(-1)` Specific conductance, (k) `=(0.367)/(36.1)=0.0116Omegacm^(-1)` Molar conductance, `Delta_(m)=(1000xxk)/(M)` `=(1000xx0.0116)/(0.08)=232ohm^(-1)cm^(2)mol^(-1)`

Discussion

91936.	6 g urea is dissolved in 90 g water. The relative lowering of vapour pressure is equal to:
Answer» 0.0196 0.06 1.1 0.02 Answer :A

Discussion

91937.	6 g of urea is dissolved in 90g of boiling water. The vapour pressure of the solution is
Answer» `744.8 MN` 758 mm 761 mm 760 mm Answer :A

Discussion

91938.	99% of a first order reaction was completed in 32 in. When wil 99.9% of the reaction complete?
Answer» 50 min 46 min 49 min 48 min Answer :D

Discussion

91939.	6 g ofH_(2) reacts with 14 g N_(2)to form NH_(3)till the reaction completely consumes the limiting reagent. The mass of other reactant (in g) left are ……
Answer» Solution :`3H_(2) + N_(2) RARR 2NH_(3)` GIVEN moles : `(6)/(2) = 3 ""(14)/(28) = 0.5` 3 mol of`H_(2)`REQUIRED 1 mol of`N_(2)`so `N_(2)`is the limiting REAGENT. 0.5 mol of `N_(2)`.will REACT with 1.5 mol`H_(2)`. So 1.5 mol`H_(2)`will be left. Mass of 1.5 mol`H_(2) = 2 xx 1.5 g = 3g`

Discussion

91940.	6 g of the organic compound on heating with NaOH gave NH_(3 which is neutralised by 200 ml of 1 NHCl. Percentage of nitrogen is :
Answer» `12 %` `60%` `46.67%` `26.67%` SOLUTION :N//A

Discussion

91941.	6 g of a substance is dissolved in 100 g of water depresses the freezing point by 0.93^(@)C. The molecular mass of the substance will be: (K_(f) for water= 1.86^(@)C/molal)
Answer» 60 g/mole 180 g/mole 120 g/mole 240 g/mole Answer :C

Discussion

91942.	6 - 8% solution of acetic acid in water is called _______.
Answer» Vinegar Glacial ACETIC acid Formalin Both (a) and (B) ANSWER :A

Discussion

91943.	99% of a 1st order reaction completed in 2.303 minutes. What is the rate constant and half life of the reaction
Answer» 2.303 and 0.3010 2 and 0.3465 2 and 0.693 0.310 and 0.693 Answer :B

Discussion

91944.	6 - amino purine is
Answer» ADENOSINE ADENINE CYTOSINE Thymine Answer :B

Discussion

91945.	""_(98)Cf^(246) was formed along with a neutron when an unknown radioactive substance was bombarded with ""_(6)C^(12). The unknown substance was
Answer» `""_(91)Pa^(234)` `""_(90)^(Th^(234)` `""_(92)U^(235)` `""_(92)U^(238)` Solution :Let the unknown substances is `""_(Z)^(A)Q` `""_(Z)^(A)Q + ""_(6)^(12)C to ""_(98)^(246)Cf + ""_(0)^(1)N` `:. Z + 6 = 98 implies Z = 92` `A + 12 = 246 + 1 implies A = 235` So, the unknown substance is `""_(92)^(235)U`.

Discussion

91946.	5ml of standard gold sol, needs 0.5 mg of gelation for its protection from coagulation, calculate the gold no of gelation.
Answer» 1 2 3 4 Solution :`5ml`of GOLD sols, required gelatin for PROTECTION from COAGULATION `=0.5mg.` `10ML` of gold sols. Required gelatin for protection from coagulation `=(0.5xx10)/(5)=1`

Discussion

91947.	0.8g of H_(2)SO_(4) is present in 2 litres of a solution. The molarityof the solution is
Answer» 0.1 M 0.04 M 0.2 M 0.01 M Solution :`M=(W xx 1000)/("mol. MASS" xx "Volume in ML.")=(9.8xx1000)/(98xx2000)=0.05 M`

Discussion

91948.	5mL of acetone is mixed with 100mL of H_(2)O. The vapour pressure of water above the solution is
Answer» <P>equal to the vapour PRESSURE of pure water equal to the vapour pressure of the solution less than the vapour pressure of pure water more than the vapour pressure of pure water Solution :`(p_(A))/(p_(A)^(@)) = x_(A)` In a solution `x_(A) lt 1` `:. p_(A) lt p_(A)^(@)`

Discussion

91949.	98% H_2SO_4 is:
Answer» Pyrosulphuric acid Oleum Azeotropic mixture None of the above Answer :C

Discussion

91950.	9.7xx10^(17) atoms of iron weigh as much as 1 cc of H_(2) at S.T.P. What is the atomic mass of iron?
Answer» SOLUTION :`"Mass of 1 cc of "H_(2)" at S.T.P."=(2.016)/(22400)=9.0xx10^(-5)g` Thus, `9.7xx10^(17)` atoms or iron have mass `=9.0xx10^(-5)g` `therefore" Mass of ONE mole, i.e., "6.02xx10^(23)" atoms of Fe"=(9.0xx10^(-5)xx6.02xx10^(23))/(9.7xx10^(17))=55.9g`

Discussion

Explore topic-wise InterviewSolutions in .

A 0.01m aqueous solution of AlCL_3 freezes at -0.068^@C Calculate the percentage of dissociation.

._(6)^(13)C and ._(7)^(14)N are the

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A 0.0020m aqueous solution of an ionic compound Co(NH_(3))_(5)(NO_(2))Clfreezes at -0.00732^(@)C. Number of moles of ions which 1 mol of ionic comound produces on being dissolved in water be (K_(f)=-1.86^(@)C//m)

A 0.0020 m aqueous solution of an ionic compound Co(NH_(3))_(5) (NO_(2))Cl freezes at -0.00744^(@)C. The number of moles of ions which 1 mole of ionic compound produces on being dissolve in water is (K_(f)=-1.86^(@)C//m)

A 0.0020 M aqueous solution of an ionic compound Co(NH_(3))_(4)(NO_(2))_(2)Cl freeze at - 0.00732^(@)C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be (K_(f)=1.86^(@)C//m).

A 0.0020 m aqueous solution in an ionic compound Co(NH_3)_5 (NO_2)Clfreezes at -0.00732^@C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be (K_f=+1.86^@ C/m)

6 grams of a substance 'A' dissolved in 100 g of water freezes at - 0.93^(@)C. Themolecular mass of 'A' is (K_(f)=1.86 Km^(-1)).

A 0.001 molal solution of a complex with molecular formula [Pt(NH_(3))_(4)Cl_(4)] in water showed a freezing point depression of 0.0054^(@)C. If K_(f) for water is 1.80, what is the correct formulation of the molecule with a proper coordination sphere?

6 gm nitrogen on successive reaction with different compounds gets finally converted into 30 gm [CR(NH_(3))_(x)Br_(2)] value of x is [ Atomic mass of Cr = 52, Br =80]

A 0-05 M sodium hydroxide offered a resistance of 31.6Omega in a conductivity cell at 298 K. Calculate the molar conductance (cell constant of cell =0367cm^(-1)).

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99% of a first order reaction was completed in 32 in. When wil 99.9% of the reaction complete?

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6 g of the organic compound on heating with NaOH gave NH_(3 which is neutralised by 200 ml of 1 NHCl. Percentage of nitrogen is :

6 g of a substance is dissolved in 100 g of water depresses the freezing point by 0.93^(@)C. The molecular mass of the substance will be: (K_(f) for water= 1.86^(@)C/molal)

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99% of a 1st order reaction completed in 2.303 minutes. What is the rate constant and half life of the reaction

6 - amino purine is

""_(98)Cf^(246) was formed along with a neutron when an unknown radioactive substance was bombarded with ""_(6)C^(12). The unknown substance was

5ml of standard gold sol, needs 0.5 mg of gelation for its protection from coagulation, calculate the gold no of gelation.

0.8g of H_(2)SO_(4) is present in 2 litres of a solution. The molarityof the solution is

5mL of acetone is mixed with 100mL of H_(2)O. The vapour pressure of water above the solution is

98% H_2SO_4 is:

9.7xx10^(17) atoms of iron weigh as much as 1 cc of H_(2) at S.T.P. What is the atomic mass of iron?