94420 + Interview Questions in Chemistry in Class 12 Page 1862 InterviewSolution

93051.	100 ml 0.2 M Na_(2)Ais titrated with 100ml 0.03 M HCl. Calculate pH of final solution. pKa_(1)(H_(2)A)=5""pKa_(2)(H_(2)A)=9
Answer» 2 7.5 5 9 Solution :`SO_(2)Cl_(2)(g)rarrSO_(2)(g)+Cl_(2)(g)` 0.5 `0.5-x ""x""x` `K_(A)=(1)/(100)LN.(0.5)/(0.25)=(ln2)/(100)` Rate `=KP_(A) = (ln2)/(100)xx0.3=2.1xx10^(-3)"atm"//"sec"`

Discussion

93052.	100 mL 0.1 M solutions of ammonium acetate is diluted by adding 100 mL of water. The pH of the resulting solution will be (pK_(a) of acetic acid is nearly equal to pK_(a) of NH_(4)OH)
Answer» 4.9 `5.0` `7.0` `10.0` Solution :Hydrolysis of a salt of weak acid and weak base `PH = 7 + (1)/(2)(pK_(a) -pK_(B)) = 7` as `pK_(a) ~~ pK_(b)`.

Discussion

93053.	100 ml, 0.05 M CuSO_4 solution is electrolysed by using current of 0.965 Å for 100 min.Find the pH of solution at the end of electrolysis.
Answer» SOLUTION :pH=1 After 1000 second when total `Cu^2+` discharge then `H^+` and `OH^(-)` discharge equally so pH will not change further. `Cu^(+2)+2e^(-)to Cu` `n_(Cu^(+2))/1=n_(E^(-))/2=Q/(2F)=(it)/(2F)` `(0.925xxt)/(2xx96500)=(0.05xx100)/1000` t=1000 sec `(n_(H^+))_("excess")=(n_(OH^(-)))_("discharge")=n_(e^(-))=Q/F=(it)/(F)=(0.965xx1000)/96500=10^(-2)` `n_(H^+)=(MV)/1000` `10^(-2)=(Mxx100)/1000implies M=10^(-1)` `[H^+]=10^(-1)` `pH= -"log" 10^(-1)=1`

Discussion

93054.	100 mg of a protein is dissolved in just enough water to make 10.0 mL of solution. If this solution has an osmotic pressure of 13.3 mm Hg at 25^(@)C, what is the molar mass of protein?
Answer» Solution :`M_(2)=(WRT)/(PIV)=((8.95xx10^(-3)g)(0.0821"L ATM K"^(-1)"mol"^(-1))(298K))/(("0.335/760 atm")(35xx10^(-3)L))="14193.3 g mol"^(-1)`

Discussion

93055.	100 grams of each P_(4)O_(6) and KMnO_(4)^(-) were mixed in hydrochloric acid solution to from H_(3)PO_(4) and MnCl_(2). Which reagentis left unreacted and how much of it is left ?
Answer» Solution :`MN^(2+) + 5E^(-) rarr Mn^(2+)` Thus milliequivalents of `KMnO_(4) = (100 xx 5 x 1000)/(158) = 3164.56` `(P^(3+))_(4) rarr 4P^(5+) + 8e^(-)` Thus MILLI equivalents of `P_(4)O_(6) = (100 xx 8 xx 1000)/(220) = 3638` Therefore `P_(4)O_(6)` is excess. Excess milliequivalents of `P_(4)O_(6) = 473.44` Weight of `P_(4)O_(6)` left unreacted `(437.44 xx 0.220)/(8) = 12.03 g`

Discussion

93056.	10.0 gram-atom of an alpha-active radioisotope is disintegrating in a sealed container. In one hour, the He gas collected at STP is 11.2 litres. Calculate the half-life of the isotope supposing each nucleus yielding one He atom.
Answer» ANSWER :13.49h

Discussion

93057.	100 gms of 118% oleum reacted with excess water. How much H_(2)SO_(4) is formed?
Answer» 90 gm 118 gm 48 gm 100 gm SOLUTION :`SO_(3)+H_(2)SO_(4) to H_(2)S_(2)O_(7), H_(2)S_(2)O_(7)+H_(2)O to 2H_(2)SO_(4)` `SO_(3)` is absorbed in `98%` conc. `H_(2)SO_(4)` to get oleum. Oleum `[H_(2)S_(2)O_(7)]` is diluted with water to get `H_(2)SO_(4)` 118% of oleum `to` 100 gm `"?" larr 100%`, 118grams.

Discussion

93058.	1.00 gm of a non-electrolyte solute dissolved in 50 gm of benzene lowered the freezing point of benzene by 0.40 K. K_(f) for benzene is 5.12 kg mol^(-1). Molecular mass of the solute will be
Answer» `256 g mol^(-1)` `2.56 g mol^(-1)` `512xx10^(3)g mol^(-1)` `2.56xx10^(4)g mol^(-1)` Solution :By USING, `m=(K_(f)xx1000xx W)/(DELTA T_(f)xx W_("SOLVENT")(gm))=(5.12xx1000xx1)/(0.40xx50)` = 256 gm/mol Hence, molecular mass of the SOLUTE `= 256 gm mol^(-1)`

Discussion

93059.	100 g of sucrose solution in water is colled to -0.5^(@)C. What weight of ice would be separated out at this temperature if solution started to freeze at -0.38^(@)C? K_(f) for H_(2)O="1.86 K kg mol"^(-1).
Answer» Solution :Suppose 100 g of the solution contains `w_(2)` g of the SOLUTE and `w_(1)` g of the solvent. Then `w_(1)+w_(2)=100"…(i)"` `DeltaT_(f)=(1000xxK_(f)xxw_(2))/(w_(1)xxM_(2))` As the solution starts freezing at `-0.38^(@)C`. Hence, when `DeltaT_(f)=0.38^(@)` `0.38=(1000xx1.86xxw_(2))/(w_(1)xx342)"or"(w_(2))/(w_(1))=0.07"...(ii)"` Solving eqns. (i) and (ii), we get `w_(2)=6.6g, w_(1)=93.40 g` Now at `-0.5^(@)C`, some WATER SEPARATES out as ice but solute exists as such, i.e., `w_(2)=6.6g`. Let us caluclate water present `(w_(1))` `0.50=(1000xx1.86xx6.6)/(w_(1)xx342)"or"w_(1)=71.78` `therefore"Weight of iceseparated out "=93.40-71.78=21.62g`

Discussion

93060.	100 g of liquid A (molar mass 140 g mol^(-1)) was dissolved in 1000 g of liuquid B (molar mass 180 g mol^(-1)). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.
Answer» Solution :NUMBER of moles of liquid A : `n_(A)=(100)/(140)=0.7143` mol Number of moles of liquid B : `n_(B)=(1000)/(180)=5.556` mol Then mole fraction of A, `x_(A)=(n_(A))/(n_(A)+n_(B))` `= (0.714)/(0.714+5.556)=0.114` And, mole fraction of B, `= x_(B)=1-0.114` = 0.8861 VAPOUR pressure of pure liquid B, `p_(B)^(0)=500` TORR Therefore, vapour presure of liquid B in the solution, `p_(B)=p_(B)^(0)x_(B)=500xx0.886` = 443 torr Total vapour pressure of the solution of liquid A in hte solution, `p_(A)=p_("total")-p_(B)` `= 475-433` = 32 torr Now,`p_(A)^(0)x_(A)` `p_(A)^(0)=(p_(A))/(x_(A))` `= (32)/(0.114)=280.5` torr Hence, the vapour pressure of pure liquid A is 280.7 torr.

Discussion

93061.	100 g of liquid A (molar mass "140g mol"^(-1)) was dissolved in 1000 g of liquid B (molar mass "180 g mol"^(-1)). The vapour pressure of pure liquid B was found o be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 torr.
Answer» Solution :`"No. of moles of liquid A (solute)"=("100 G")/("140 g mol"^(-1))=(5)/(7)"MOLE"` `"No. of moles of lliquid B (solvent)"=("1000 g")/("180 g mol"^(-1))=(50)/(9)" mole"` `THEREFORE"Mole fraction of A in the solution "(x_(A))=(5//7)/(5//7+50//9)=(5//7)/(395//63)=(5)/(7)xx(63)/(395)=(45)/(395)=0.114` `therefore"Mole fraction of B in the solution "(x_(B))=1-0.114=0.886` `"Also, given"p_(B)^(@)="500 torr"` `"Applying Raoult's law,"p_(A)=x_(A)p_(A)^(@)=0.114xxp_(A)^(@)"...(i)"` `p_(B)=x_(B)p_(B)^(@)=0.886xx500="443 torr"` `p_("Total")=p_(A)+p_(B)` `475=0.114p_(A)^(@)+443"or"p_(A)^(@)=(475-443)/(0.114)="280.7 torr"` Substituting this value in eqn. (i) we get `""p_(A)=0.114xx"280.7 torr = 32 torr."`

Discussion

93062.	100 g of liquid A (molar mass 140 g "mol"^(-1) ) was dissolved in 1000 g of liquid B (molar mass 180 g "mol"^(-1) ). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 torr.
Answer» Solution : Number of moles of liquid A (solute) = `100/140 = 5/7 `mole Number of moles of liquid B (solvent) = `1000/180 = 50/9 `moles ` THEREFORE ` Mole fraction of A in the solution `(x_A) =(5//7)/(5//7 + 50//9) = (5//7)/(395//63) = 5/7 XX 63/395 = 45/395 = 0.114` ` therefore `Mole fraction of B in the solution (`x_B`) = 1 – 0.114 = 0.886 Also, given `p_B^0` = 500 torr Applying Raoult.s law and substituting the values, we have `p_A =x_A p_A^0 = 0.114 xx p_A^0` `p_B = x_B p_B^0 = 0.886 xx 500 = 443`torr `p_("total") =p_A + p_B` `475 = 0.114 p_A^0 + 443` `p_A^0 = (475 - 443)/(0.114) = 280.7` torr Vapour PRESSURE of A in the solution `p_A = 0.114 xx 280.7` torr = 32 torr.

Discussion

93063.	100 g of argon is allowed to expand from a pressure of 10 atm to 0.1 atm at 100^(@)C . Calculate the heat which is obsorbed, assuming ideal behaviour.
Answer» ANSWER :(8580 CAL)

Discussion

93064.	100 g of a sample of HClsolution of relative density 1.17 contains 33.4 g of HCl . What volume of this HCl solution will be required to neutralise exactly5 litres of N/10NaOH solution ?
Answer» SOLUTION :Volume of HCl solution ` = 100/(1.17) mL ` `(" density " = ("mass")/("volume"))` Equivalents of HCl = `(33.4)/(36.5)` (eq.wt . Of HCl = 36.5) m.e of HCl ` = (33.4)/(36.5) xx 1000 "" … (Eqn . 3)` NORMALITY of HCl = `("m.e")/(" volume in mL") "" ....(Eqn .1)` ` =(33.4)/(36.5) xx 1000 xx (1.17)/100` ` = 10.7 ` N Now let the volumeof HCl, of normality calculated above , REQUIRED to neutralise exactly the given NaOH solution be v mL . Now let the volume of HCl , of normality calculated above , required to neutralise exactly the given ,naOH solution be v mL . m.e of HCl = m.e of NaOH `10.7 xx v = 1/10 xx 5000` ` 10 . 7xx v 1/10 xx 5000` `10.7 xx v = 500` ` :. "" v = 46.7 mL `

Discussion

93065.	1.00 g of a non-electrolyte solute (molar mass 250g mol^(-1) was dissolved in 51.2 g of benzene. If the freezing point depression constant, K_(f) of benzene is 5.12 kh mol^(-1), the freezing point of benzene will be lowered by
Answer» 0.5 K 0.2 K 0.4 K 0.3 K SOLUTION :`DELTA T_(f)=(K_(f)XX W_(B)xx1000)/(M_(B)xx W_(A))` `Delta T_(f)=(5.12xx1.0xx1000)/(250xx51.2)` `Delta T_(f)=0.4 K`

Discussion

93066.	1.00 g of a non electrolyte solute ( molar mass 250 " g mol"^(-1)) was dissolved in 51.2 g of benzene. If the freezing point depression constant, K_(f) of benzene is 5*12 K kg "mol"^(-1), the freezing point of benzene will be lowered by
Answer» 0.3 K 0.5 K 0.2 K 0.4 K Solution :`DELTA T_(f)=(K_(f)xx 1000 xx W_(2))/(M_(2)xxW_(1))` `(512xx1000xx1)/(512xx250)=0*4`.

Discussion

93067.	1.00 g of a non - electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constnat of benzene is 5.12 K kg mol^(-1). Find the molar mass of the solute.
Answer» Solution :SUBSTITUTING the VALUES of various terms INVOLVED in following EQUATION, we get `M_(2)=(K_(f)xx w_(2)xx 1000)/(Delta T_(f)xx w_(1))` `M_(2)=(5.12"K kg mol"^(-1)xx1.00g xx 1000 g kg^(-1))/(0.40xx50 g)` `= 256g mol^(-1)` THUS, molar mass of the solute `= 256 g mol^(-1)`

Discussion

93068.	1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is "5.12 K kg mol"^(-1). Find the molar mass of solute.
Answer» SOLUTION :Here, we are given `w_(2)=1.00 g, w_(1)=50, DeltaT_(f)=0.40 K, K_(f)="5.12 K kg MOL"^(-1)` Substituting these values in the formula `M_(2)=(1000K_(f)w_(2))/(w_(1)DeltaT_(f)),` we get `M_(2)=("1000 g kg"^(-1)xx"5.12 kg mol"^(-1)xx1.0g)/(50 g xx0.40 K)="256 g mol"^(-1)`

Discussion

93069.	100 cm^(3) of NH_(3) diffuses through a fine hole in 32.5 seconds. How much time will 60 cc of N_(2) take to diffuse under the same conditions ?
Answer» ANSWER :25 s

Discussion

93070.	100 cm^(3) of a given sample of H_(2)O_(2) gives 1000 cm^(3) of O_(2) at S.T.P. The given sample is
Answer» 10 VOLUME `H_(2)O_(2)` 90% of `H_(2)O_(2)` 10% of `H_(2)O_(2)` 100 volume `H_(2)O_(2)` Solution :`underset("100 ML")(2H_(2)O_(2))rarr 2H_(2)O+underset("1000 mL")(O_(2))` or 1mL of `H_(2)O_(2)` will give 10 mL of `O_(2)` of STP. Thus its volume strengthis 10 volume.

Discussion

93071.	100 cm^(3) of 1M CH_(3)COOH and 100 cm^(3) of 2 M CH_(3)OH were mixed to form an ester. The change in their intial rate if each solution is mixed with equal volume of water would be:
Answer» 4 TIMES `0.25` times 2 times `0.5` times. SOLUTION :N//A

Discussion

93072.	100" cm"^(3)" of "1" M "CH_(3)COOH was mixed with 100" cm"^(3)" of "2" M "CH_(3)OH to form an ester. The change in the initial rate if each solution is diluted with equal volume of water would be
Answer» 4 times 0.25 times 2 times 0.5 times Solution :`underset(1M)(CH_(3)COOH)+underset(2M)(CH_(3)OH)toCH_(3)COOCH_(3)+H_(2)O` Rate `(r_(1)) = K[CH_(3)COOH][CH_(3)COH]` `= k(1)(2) = 2K` When each solution is DILUTED with equal volume of WATER, concentration of each is halved, Now, rate `r_(2) = k((1)/(2))((2)/(2)) =(k)/(2) therefore (r_(2))/(r_(1)) = (k//2)/(2k) = (1)/(4) = 0.25` or `r_(2) =0.25 r_(1)`

Discussion

93073.	100 cm^(3) of 0.1 N HCl solution is mixed with 100 cm^(3) of 0.2 N NaOH solution. The resulting solution is
Answer» 0.1 N and the solution is basic 0.05 N and the solution is basic 0.1 N and the solution is acidic 0.05 N AD the solution is acidic Solution :Acid=0.1`xx100=10` base`=0.2xx100=20` `THEREFORE` Solution will be basic. 0.1 N of HCl is NEUTRALISED by 0.1 N of NaOH and the remaining 0.1 N of NaOH is in 200 `cm^(3)` of solution. `therefore` Resulting normality`=0.5`N, basic.

Discussion

93074.	100 cm^3 of 1 M CH_3COOH was mixed with 100 cm^3 of 2 M CH_3OH to form an ester. The change in the initial rate if eachsolution is diliuted with equal volume of water would be
Answer» 0.5 TIMES 2 times 0.25 times 4 times ANSWER :C

Discussion

93075.	10% (w/v) solution of sucrose (M.W = 342) is isotonic with 5% (w/v) solution of a solute X under identical conditions. The molecular weight of X is
Answer» 171 242 684 342 Answer :A

Discussion

93076.	10 volumes of H_2O_2 has a strength of approximately:
Answer» `3%` `30%` `10%` `5%` ANSWER :A

Discussion

93077.	10 volume of H_(2)O_(2) = x(N)H_(2)O_(2). What is the value of x?
Answer» ANSWER :X= 56

Discussion

93078.	10% solution of urea is isotonic with 6% solution of a non-volatile solute X. What is the molecular mass of solute X?
Answer» `"6 G MOL"^(-)` `"60 g mol"^(-1)` `"36 g mol"^(-1)` `"32 g mol"^(-1)` Answer :C

Discussion

93079.	10% solution of phenol is used as a
Answer» ANTISEPTIC DISINFECTANT Both None Answer :B

Discussion

93080.	10% sites of catalyst bed have adsorbed H_2. On heating, H_2 gas is evolved from sites and collected at 0.03 atm and 300K in a small vessel of 2.46 cm^3. Number of sites available is 7.2×(10)^(16) per cm^2 and surface area is 1000cm^2. The number of surface sites are occupied by one molecule of H_2 is [Given: N_A=6×(10)^(23)]
Answer» 1 2 3 4 Answer :D

Discussion

93081.	10% sites of catalyst bed have adsorbed by H_2 on heating H_2 gas is evolved from sites and collected at 0.03atm and 300K in a small vessel of 2.46cm^3. No.of sites available is 5.4 xx 10^16 per cm^2 and surface area is 1000 cm^2. Findout the no.of surface sites occupied per molecule of H_2 (N_A = 6 x 10^23)
Answer» 1 2 3 4 Solution :1 Adsorbed moles of `H_2 = (0.03 xx 2.46 xx 10^(-3))/(0.0821 xx 300) = 3 xx 10^(-6)` No. of ABSORBED molecules of `H_2 = 3 xx 10^(-6) xx 6 xx 10^(23) = 18 xx 10^(17)` Total no. of SURFACE sites available `= 5.4 xx 10^(16) xx 1000 = 5.4 xx 10^(19) cm^2` No. of surface sites that is occupied by adsorption of `H_2 = 10/100 xx 5.4 xx 10^(19) = 5.4 xx 10^(18)` No. of surface sites occupied by one molecule `H_2 = (5.4 xx 10^(18))/(18 xx 10^(17)) = 3`.

Discussion

93082.	10% of a reactant decomposes in 1 hour, 20% in 2 hours and 30% in 3 hours. The order of the reaction is
Answer» 0 1 2 3 Answer :A

Discussion

93083.	1^(0) -nitroalkane react with HNO_(2)gives
Answer» dinitroalkane NITROSO nitroalkane N-alkyl hydroxyl AMINE nitrosoamine ANSWER :B

Discussion

93084.	10 moles of X, 12 mole of Y and 20 moles of Z are mixed to produce a final product P, according to the given balanced reaction - X + 2Y rarr I I + Z rarr Y + P then the maximum moles of P, which can be produced assuming that the products formed can also be reused in the reaction ?
Answer» 6 MOLE 9 mole 10 mole : 2 mole Solution :`{:(,X,+,2Y,rarr,I,,I,+,Z rarr,Y,+,P,),("step - I",10,,12,,,"step - I",6,,20,,,,),(,10 - 6,,,,6,,0,,20 - 6,6,,6,),("step - II",4,,6,,0,"step - II",3,,14,0,,6,),(,4 - 3,,0,,3,,0,,14 - 3,3,,6 + 3,),("step - III",1,,3,,0,"step - III",1,,11,0,,9,),(,0,,3 - 2,,1,,0,,11 -1,1,,9 + 1,):}` `= 10`MOLES of P

Discussion

93085.	10 moles of a liquid L are 50% converted into its vapour at its boiling point (273^@C) and at a pressure of 1 atm.If the value of latent heat of vapourisation of liquid L is 273 L atm/mole , than which of the following statements is/are correct : Assume volume of liquid to be negligible and vapour of the liquid to behave ideally.
Answer» Work done by the system in the above process is 224 L atm The enthalpy change `(DELTAH)` for the above process is 1365 L atm (with respect to magnitude only) The ENTROPY of the system increases by 2.5 L atm in the above process The value of `DeltaU` for the above process is 1589 L atm. Solution :`V_f=5xxRxx546=224 L` `W=-P_(ext)(DeltaV)=-1` atm (224 L) =-224 L atm `:.` work done by system =224 L atm Enthalpy change `(DeltaH)=q=273xx5=1365 L atm` `DeltaS=(DeltaH_(VAP))/T=1365/546=2.5` L atm /K `DeltaU=q+W=1365-224=1141 L atm`

Discussion

93086.	10 mole of liquid 'A' and 20 mole of liquid 'B' is mixed in a cylindrical vessel containing a piston arrangement. Initially a pressure of 2 atm is maintained on the solution. Now, the piston is raised slowely and isothermally. Assume ideal behaviour of solution and A and B are completely miscible. P_(A)^(@) = 0.6 atm and P_(B)^(@) = 0.9 atm The pressure at which (1)/(3) theof the total amount (by mol) of liquid solution taken initially will be present in the vapour form is [Given: sqrt(57) = 7.55]
Answer» `0.783` `0.800` `0.600` `0.652` ANSWER :A

Discussion

93087.	10 mole of liquid 'A' and 20 mole of liquid 'B' is mixed in a cylindrical vessel containing a piston arrangement. Initially a pressure of 2 atm is maintained on the solution. Now, the piston is raised slowely and isothermally. Assume ideal behaviour of solution and A and B are completely miscible. P_(A)^(@) = 0.6 atm and P_(B)^(@) = 0.9 atm The minimum pressure for the existence of liquid solution is
Answer» `0.6atm` `0.8atm` `0.77atm` `0.9atm` ANSWER :C

Discussion

93088.	10 mole of liquid 'A' and 20 mole of liquid 'B' is mixed in a cylindrical vessel containing a piston arrangement. Initially a pressure of 2 atm is maintained on the solution. Now, the piston is raised slowely and isothermally. Assume ideal behaviour of solution and A and B are completely miscible. P_(A)^(@) = 0.6 atm and P_(B)^(@) = 0.9 atm The pressure below which the evaporation of liquid solution will start, is
Answer» `0.5 ATM` `0.8atm` `0.77atm` `0.9atm` ANSWER :B

Discussion

93089.	10 mole of liquid 'A' and 10 mole of liquid 'B' is mixed in a cylindrical vessel containing a piston arrangement. Initially a pressure of 2 atm is maintained on the solution. Now, the piston is raised slowly and isothermally. Assuming A and B tobe completely miscible and forming an ideal solution. P_(A)^(@) = 0.6atm and P_(B)^(@) = 0.9 atm The minimum pressure for the existance of liquid solution is :
Answer» `0.6` ATM `0.8` atm `0.72` atm `0.9` atm Answer :C

Discussion

93090.	10 mole of liquid 'A' and 10 mole of liquid 'B' is mixed in a cylindrical vessel containing a piston arrangement. Initially a pressure of 2 atm is maintained on the solution. Now, the piston is raised slowly and isothermally. Assuming A and B tobe completely miscible and forming an ideal solution. P_(A)^(@) = 0.6atm and P_(B)^(@) = 0.9 atm If 1/4 of the total amount of liquid solution taken initially is converted to vapour, then moles of A in the vapour will be : [Given : sqrt(2425) = 49.25]
Answer» `2.125` `3.875` `4` `1` ANSWER :A

Discussion

93091.	10 mol sampleof AgNO_(3)is dissolved in one lit of 1.00 M NH_(3)Is it possibleAgCl(s)form the solution by adding0.010molof NaCl ?
Answer» Solution :`K_(SP(AgCl)) = 1.8 xx10^(-10), K _(F[Ag(NH_(3))_(2)^(.)])= 1.6 xx10^(7)` ` {:("Ag"^(+),+,2NH_(3),hArr,[Ag(NH_(3))_(2)^(+)]),(0.10 M ,,1.00,,0),(x,,(1-0.20)M,,),(,,=0.80 M,,0.10 M ):}` It is assumed that all `Ag^(+)` ions have been complexed and only x amount is left `K_(f)= ([Ag(NH_(3))_(2)])/([Ag^(+)][NH_(3)]^(2)) RARR 1.6 xx10^(7) = (0.10)/(XX(0.80)^(2))` ` :. x = 9.8 xx10^(-9) M = [Ag^(+)] ` undissolved `[Cl^(-)] = 1.0 xx10^(-2) M ` ` :. [Ag^(+)] [Cl^(-)] = 9.8 xx10^(-9) xx1.0 xx10^(2) = 9.8 xx10^(-11) lt 1.8 xx10^(-10) [K_(sp(AgCl))] ` Hence , AgCl (s) will not precipitate

Discussion

93092.	10 mol of an ideal gas (gamma= 1.5)expands adiabatically from (400K,10L) to 20 L along following paths. Path A : Reversible Path B : One step expansion. Path C : Two step expansion. Path D : Free expansion Which is/are correct statement(s) ?
Answer» Final temperature attained along path B is maximum. Total change in entropy along path-D is 10R In2. Change in entropy of gas along path A is zero. Final temperature attained along path D is 400 K. Solution :`DeltaU=q+w=o +w=w` `n_(Cvm)(T_(2)-T_(1))=w` `T_(2)=T_(1)+(w)/(n_(Cvm))` `w=-x` `T_(2)=T_(1)+(x)/(n_(Cvm))` x is minimum along path B. So `T_(2)` is maximum. Pathd-D `DeltaS_("total")=DeltaS_("SYSTEM")+DeltaS_("SURROUNDING")` `Delta_("total")=nC_(vm)ln .(T_(2))/(T_(1))+nRln.(V_(2))/(V_(1))+O` `T_(2)=T_(1)` (for free expansion) `DeltaS_("total")=10Rln2` Path A `DeltaS_("gas")=0` for reversible adiabatic process. Path-D `T_(2)=T_(1)+(w)/(n_(Cvm))w=0` `T_(2)=T_(1)=400K`

Discussion

93093.	10 mL of tap water containing ca^(2+) and Mg^(2+) in the presence of HCO_(3)^(-)was properly buffered and the indicator mureside added . The sample was dilutedand heated to 60^(@)C.Titration with 0.01M EDTA solution changed the indicator colour at 7.50 mL . This complexed Ca^(2+) only . A second 10.mL sample was made basic and Erio T indicator added . Titration with 0.01M EDTAsolutionchanged the indicator colourat 13.02 mLUndertheseconditions both ca^(2+) and Mg^(2+) are complexed . If the 10 mLof water samplewere to be evaporated to dryness , what weight of CaCO_(3)+ MgCO_(3) would be formed?
Answer» Solution :All EDTA cmplxesare FORMED on a ONE to one basis with dipositive ions . Mole f `Ca^(2+) +MG^(2+) ` = mole of `CaCO_(3)` + mole of `MgCO_(3)` `= (0.01 xx13.02)/(1000) = 13 xx 10^(-5)` Mole of `Ca^(2+) ` mole of `CaCO_(3)` `= (0.01 xx 7.50)/1000 = 7.50 xx10^(-5)` ` :. ` mole of `MgCO_(3) = 13 xx 10^(-5) - 7.50 xx10^(-5) = 5.5 xx 10^(-6)` ` :. `weight of `CaCO_(3) +MgCO_(3)= 7.50 xx10^(-5) xxx100 + 5.5 xx10^(-5) xx 84` ` = 1.21 xx 10^(-2) g ` `(CaCO_(3) = 100 , MgCO_(3)=84)`

Discussion

93094.	10ml of sulphuric acid solution (sp.gr.=1.84) contains 98% weight of pure acid. Calculate the volume of 2.5 M NaOH solution required to just neutralise the acid.
Answer» SOLUTION :`147.2` ML

Discussion

93095.	10 mL of mixture containing carbon monoxide and nitrogen required 7mL, oxygen to form CO_(2) and NO, on combustion The volume of N_(2) in the mixture will be:
Answer» 7/2mL 17/2mL 4mL 7mL Solution :`CO(g)+(1)/(2)O_(2)(g)rarrCO_(2)(g)` `N_(2)(g)+O_(2)(g)rarr2NO(g)` `X+y=10` `(x)/(2)+y=7` Solving eqs, i and II, `x=6` and y=4

Discussion

93096.	10mlof NaHC_(2)O_(4) is oxidized by 10 ml of 0.02 M MnO_(4)^(-) . Therefore, 10 ml of NaHC_(2)O_(4) can be neutralized by
Answer» 10 ML of `0.1` M NaOH 10 ml of `0.02` MNaOH `10 ml" of " 0.1 N CA (CaOH)_(2)` 10 mlof`0.05 MBA(OH)_(2)` SOLUTION :N/A

Discussion

93097.	10ml of liquid 'A' is mixed with 10ml of liquid 'B' the volume of the resultant solution is 19.9 ml. What type of deviation expected from Raoult's law?
Answer» SOLUTION :NEGATIVE DEVIATION.

Discussion

93098.	10 mL of liquid carbon disulphide (sp. Gravity 2.63) is burnt in oxygen. Find the volume of the resulting gases measured at S.T.P.
Answer» Solution :`"10 mL of CS"_(2)=10xx2.63g=26.3g` `{:(UNDERSET(76g)(CS_(2))+3O_(2)rarr underset("3 moles")(ubrace(CO_(2)+2SO_(2)))),(""=3xx22.4" L at STP"):}`

Discussion

93099.	10 mL of liquid 'A' is mixed with 10 mL of liquid 'B', the volume of the resultant solution is 19.9 ml. What type of deviation expected from Raoult's law ?
Answer» SOLUTION :NEGATIVE DEVIATION from Raoul.s LAW or -ve deviation from Raoult.s law.

Discussion

93100.	10 mL of HCl solution gave 0.1435 g of AgCl when treated with excessof AgNO_(3). Find the normality of the acid solution [Ag = 108]
Answer» SOLUTION :`"CL present in 0.1435 g AgCl"=(35.5)/(143.5)xx0.1435g=0.0355g` `"HCL containing 0.0355 g Cl"=(36.5)/(35.5)xx0.0355g=0.0365g=(0.0365)/(36.5)"g eq."="0.001 g eq."`

Discussion

Explore topic-wise InterviewSolutions in .

100 ml 0.2 M Na_(2)Ais titrated with 100ml 0.03 M HCl. Calculate pH of final solution. pKa_(1)(H_(2)A)=5""pKa_(2)(H_(2)A)=9

100 mL 0.1 M solutions of ammonium acetate is diluted by adding 100 mL of water. The pH of the resulting solution will be (pK_(a) of acetic acid is nearly equal to pK_(a) of NH_(4)OH)

100 ml, 0.05 M CuSO_4 solution is electrolysed by using current of 0.965 Å for 100 min.Find the pH of solution at the end of electrolysis.

100 mg of a protein is dissolved in just enough water to make 10.0 mL of solution. If this solution has an osmotic pressure of 13.3 mm Hg at 25^(@)C, what is the molar mass of protein?

100 grams of each P_(4)O_(6) and KMnO_(4)^(-) were mixed in hydrochloric acid solution to from H_(3)PO_(4) and MnCl_(2). Which reagentis left unreacted and how much of it is left ?

10.0 gram-atom of an alpha-active radioisotope is disintegrating in a sealed container. In one hour, the He gas collected at STP is 11.2 litres. Calculate the half-life of the isotope supposing each nucleus yielding one He atom.

100 gms of 118% oleum reacted with excess water. How much H_(2)SO_(4) is formed?

1.00 gm of a non-electrolyte solute dissolved in 50 gm of benzene lowered the freezing point of benzene by 0.40 K. K_(f) for benzene is 5.12 kg mol^(-1). Molecular mass of the solute will be

100 g of sucrose solution in water is colled to -0.5^(@)C. What weight of ice would be separated out at this temperature if solution started to freeze at -0.38^(@)C? K_(f) for H_(2)O="1.86 K kg mol"^(-1).

100 g of argon is allowed to expand from a pressure of 10 atm to 0.1 atm at 100^(@)C . Calculate the heat which is obsorbed, assuming ideal behaviour.

100 g of a sample of HClsolution of relative density 1.17 contains 33.4 g of HCl . What volume of this HCl solution will be required to neutralise exactly5 litres of N/10NaOH solution ?

1.00 g of a non-electrolyte solute (molar mass 250g mol^(-1) was dissolved in 51.2 g of benzene. If the freezing point depression constant, K_(f) of benzene is 5.12 kh mol^(-1), the freezing point of benzene will be lowered by

1.00 g of a non electrolyte solute ( molar mass 250 " g mol"^(-1)) was dissolved in 51.2 g of benzene. If the freezing point depression constant, K_(f) of benzene is 5*12 K kg "mol"^(-1), the freezing point of benzene will be lowered by

1.00 g of a non - electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constnat of benzene is 5.12 K kg mol^(-1). Find the molar mass of the solute.

1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is "5.12 K kg mol"^(-1). Find the molar mass of solute.

100 cm^(3) of NH_(3) diffuses through a fine hole in 32.5 seconds. How much time will 60 cc of N_(2) take to diffuse under the same conditions ?

100 cm^(3) of a given sample of H_(2)O_(2) gives 1000 cm^(3) of O_(2) at S.T.P. The given sample is

100 cm^(3) of 1M CH_(3)COOH and 100 cm^(3) of 2 M CH_(3)OH were mixed to form an ester. The change in their intial rate if each solution is mixed with equal volume of water would be:

100" cm"^(3)" of "1" M "CH_(3)COOH was mixed with 100" cm"^(3)" of "2" M "CH_(3)OH to form an ester. The change in the initial rate if each solution is diluted with equal volume of water would be

100 cm^(3) of 0.1 N HCl solution is mixed with 100 cm^(3) of 0.2 N NaOH solution. The resulting solution is

100 cm^3 of 1 M CH_3COOH was mixed with 100 cm^3 of 2 M CH_3OH to form an ester. The change in the initial rate if eachsolution is diliuted with equal volume of water would be

10% (w/v) solution of sucrose (M.W = 342) is isotonic with 5% (w/v) solution of a solute X under identical conditions. The molecular weight of X is

10 volumes of H_2O_2 has a strength of approximately:

10 volume of H_(2)O_(2) = x(N)H_(2)O_(2). What is the value of x?

10% solution of urea is isotonic with 6% solution of a non-volatile solute X. What is the molecular mass of solute X?

10% solution of phenol is used as a

10% of a reactant decomposes in 1 hour, 20% in 2 hours and 30% in 3 hours. The order of the reaction is

1^(0) -nitroalkane react with HNO_(2)gives

10 moles of X, 12 mole of Y and 20 moles of Z are mixed to produce a final product P, according to the given balanced reaction - X + 2Y rarr I I + Z rarr Y + P then the maximum moles of P, which can be produced assuming that the products formed can also be reused in the reaction ?

10 mol sampleof AgNO_(3)is dissolved in one lit of 1.00 M NH_(3)Is it possibleAgCl(s)form the solution by adding0.010molof NaCl ?

10 mol of an ideal gas (gamma= 1.5)expands adiabatically from (400K,10L) to 20 L along following paths. Path A : Reversible Path B : One step expansion. Path C : Two step expansion. Path D : Free expansion Which is/are correct statement(s) ?

10ml of sulphuric acid solution (sp.gr.=1.84) contains 98% weight of pure acid. Calculate the volume of 2.5 M NaOH solution required to just neutralise the acid.

10 mL of mixture containing carbon monoxide and nitrogen required 7mL, oxygen to form CO_(2) and NO, on combustion The volume of N_(2) in the mixture will be:

10mlof NaHC_(2)O_(4) is oxidized by 10 ml of 0.02 M MnO_(4)^(-) . Therefore, 10 ml of NaHC_(2)O_(4) can be neutralized by

10ml of liquid 'A' is mixed with 10ml of liquid 'B' the volume of the resultant solution is 19.9 ml. What type of deviation expected from Raoult's law?

10 mL of liquid carbon disulphide (sp. Gravity 2.63) is burnt in oxygen. Find the volume of the resulting gases measured at S.T.P.

10 mL of liquid 'A' is mixed with 10 mL of liquid 'B', the volume of the resultant solution is 19.9 ml. What type of deviation expected from Raoult's law ?

10 mL of HCl solution gave 0.1435 g of AgCl when treated with excessof AgNO_(3). Find the normality of the acid solution [Ag = 108]