94420 + Interview Questions in Chemistry in Class 12 Page 1742 InterviewSolution

87051.	A solution when dilutedwith H_(2)O And bolled gives a white precipitate .On the additionof excess NH_(2)CINH_(4)OH the volumeof theprecipitatedecreases leavingg bebinda whitege3lationtiousprecipitateidentifytheprecipitatewhichdissolves in NH_(4)OH//NH_(4)Cl :
Answer» `ZN(OH)_(2)` `AL(OH)_(3)` `Mg(OH)_(2)` `Ca(OH)_(2)` Solution :`Zn^(2+) + 2H_(2)O rarrunderset(WHITE)(Zn(OH)_(2) DARR)+ 2H^(oplus)` `Zn(OH)_(2) + 2NH_(4)CI + 2NH_(4)OH rarr underset("soluble")([Zn(NH_(3))_(4)]CI_(2)) + 4H_(2)O` `Zn(OH)_(2)` precipitatedissolvesin excess of `NH_(4)OH` in the presence of `NH_(4)CI` toformtetrqamine solublecomplex

Discussion

87052.	A solution when diluted with H_(2)O and boiled, it gives a white precipitate. On addition of excess NH_(4)Cl//NH_(4)OH, the volume of precipitate decreases leaving behind a white gelatinous precipitate. Identify the precipitate which dissolves in NH_(4)OH//NH_(4)Cl.
Answer» `ZN(OH)_(2)` `Al(OH)_(3)` `Mg(OH)_(2)` `Ca(OH)_(2)` ANSWER :A

Discussion

87053.	A solution when diluted with H_(2)O and boiled, it gives a white precipitate. On addition of excess NH_(4)Cl//NH_(4)OH the volume of precipitate decreases leaving behind a white gelatinous precipitate. Identify the precipitate which dissolves in NH_(4)OH//NH_(4)Cl
Answer» `Zn(OH)_(2)` `Al(OH)_(3)` `Mg(OH)_(2)` `CA(OH)_(2)` SOLUTION :DUE to formation of tetraammine ZINC (II) complex, `Zn^(+2)+NH_(4)OH to [Zn(NH_(3))_(4)]^(+2)`

Discussion

87054.	A solution that obeys Raoult's law is called
Answer» NORMAL solution non-ideal solution ideal solution saturated solution Answer :C

Discussion

87055.	A solutiion that contain more solute than the saturated solution at a given termperature ________ solution .
Answer» ANSWER :C

Discussion

87056.	A solution prepared by dissolving 8.95 mg of a gene fragment in35.0 ml of water has an osmotic pressure of 0.335 ton at 25^(@)C. Assuming thegene fragment is a non-electrolyse, determine the molar mass.
Answer» Solution :Mass of gene fragment = 8.95 mg = `8.95 xx 10^(-3) g` Volume of water = `35.0 ML = 35 xx 10^(-3)` L `pi = 0.335` ton = 0.335/760 atm Temp = 25 + 273 = 298 K `pi = (W_(B) RT)/(M_(B) xx V)` `(0.335)/(760) = (8.95 xx 10^(-3) xx 0.0821 xx 298)/(M_(B) xx 35 xx 10^(-3))` `M_(B) = 141933 g MOL^(-3)`

Discussion

87057.	A solution prepared by dissolving 8.95 mg of a gene fragment in 35.0 mL of water has an osmotic pressure of 0.335 torr at 25^(@)C. Assuming that the gene fragment is a non-electrolyte, calculate its molar mass.
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

87058.	A solution prepared by dissolving 15 g of non-volatile solute in 270 g of water gave relative lowering of vapour pressure of 0.005. The molecular weight of the solute is
Answer» 324 200 225 20 Solution :`M_(2) = (W_2M_1)/(W_1) XX ((p^@)/(p^@ -p_s))` where, SYMBOLS havetheir usualmeaning , `M_(2) = (15 xx 18)/(270) xx (1)/(0.005) =200`

Discussion

87059.	A solution prepared by dissolving 1.25 g of oil of winter green (methyl salicylate) in 99.0 g of benzene has a boiling point of 80.31^(@)C. Determine the molar mass of this compound (B.P. of pure benzene = 80.10^(@)C and K_(b) for benzene =2.53^(@)"C kg mol"^(-1).
Answer» SOLUTION :`"152 G MOL"^(-1)`

Discussion

87060.	A solution prepared by dissolving 1.25 g of oil of winter green (methyl s.ilicylate) in 99.0 g of benzene has a boiling po . int of 80.31^(@) C , Determine the molar mass of this compound. (B.P. of pure benzene = 80.10^(@)C and K_(b) for benzene=2.53^(@) C " Kg " mol^(-1) )
Answer» Solution :`Delta T_(B) = (80.31 = 80.10)^(@)` C = `0.21^(@) C or 0.21` K `Delta T_(b) = K_(b)` m `0.21^(@) C = 2.53^(@)`CKG `MOL^(-1) XX (1.25 g)/(M) xx (1000)/(99 kg)` M = 152 g `mol^(-1)` Where M is molar mass of the SOLUTE.

Discussion

87061.	A solution prepared by dissolving 1.25g of oil of winter green (methyl sallicylate) in 99.0g of benzene has a boiling point of 80.31^(@)C. Determine the molar mass of this compound. (B.P. of pure benzene =80.10^(@)C and K_(b) for benzene =2.53^(@)C kg "mol".1)
Answer» Solution :`M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xxW_(A))` `K_(b)=2.53 K KG mol^(-1),W_(B)=1.25g, W_(A)=99.0g=0.099 kg, DeltaT_(b)=(80.31-80.10)=0.21^(@)=0.21K` `M_(B)=((2.53K kgmol^(-1))xx(1.25g))/((0.21K)xx(0.099 kg))=152.12 g mol^(-1)`

Discussion

87062.	A solution prepared by dissolving 0.8gm of naphthalene in 100g of C Cl_4 has a boiling point elevation of 0.4^@C. A 1.24 g of an un known solute in same amount of C Cl_4 produced boiling point elevation of 0.62^@C, then molar mass of un-known solute is
Answer» 25g 50g 75g 128g Answer :D

Discussion

87063.	A solution of white crystals with a solubleof Na_(2)CO_(3) .The action of cone H_(2)SO_(4) on the crystals yieds a brown gas .The crystalsare of
Answer» `NaNO_(3)` `KCI` `Ca(ON_(3))_(2)` `NaBr` Answer :d

Discussion

87064.	A solution on treatment with NH_(3) turns blue, contains
Answer» `Cu^(+2)` `NI^(+2)` `Co^(+2)` `MN^(+2)` Solution :`Cu^(2+)+4NH_(3) to UNDERSET("Deepblue colour")([Cu(NH_(3))_(4)]^(2+))`

Discussion

87065.	A solution of white crystals gives a precipitate with AgNO_(3) but no precipitate with a solution of Na_(2)CO_(3). The action of conc. H_(2)SO_(4) on the crystals yields a brown gas. The crystals are of :
Answer» `NaNO_(3)` KCL `CA(NO_(3))_(2)` NaBr Answer :D

Discussion

87066.	A solutionof weakbase BOHwas titrated with 0.1 N HCl . The pHof the solution was foundto be10.24and 9.14 after the additionof 5 ml and20ml of the acidrespectively. Findthe dissociation constantof the base .
Answer» SOLUTION :`{:("Case-I",BOH,+,HCl,to,BCl,+,H_(2)O),("Millimole before reaction",a,,0.1 xx5 = 0.5,,0,,0),("millimole after reaction ",(a-0.5),,0,,0.5,,0.5):}` ` :. pOH= - log K_(b) + log. ([BCl])/([BOH])""` …..(i) ` :.pH = 10.04` ` :.3.96 = - logK_(b) + log. (0.5)/((a-0.5))""` ....(ii) ` :. pOH= 3.96 ` Case - II :`BOH + HClto BCl +H_(2)O ` ` {:("Millimole before reaction ",a,0.1xx20=2,,),("Millimole after reaction ",(a-2),0,2,2):}` ` :. pOH = - log K_(b) + log. ([BCl])/([BOH])""` ...(III) ` :.pH = 9.14 :. pOH = 4.86 ` ` :. 4.86= - log K_(b) + log . 2/((a-2))""` ....(iv) Solving Eqs.(ii) and (iv) , `K_(h) = 1.81 xx10^(-5)`

Discussion

87067.	A solution of weak acids is diluted by adding an equal volume of water. Which of the following will not change
Answer» Strength of the acid The VALUE of `[H_(3)O^(+)]` PH of the solution The degree of dissociation of acid Solution :The value of `H_(3)O^(+)` ions will not changed. `CH_(3)COOH + H_(2)O CH_(3)COO^(-) + H_(3)O^(-)`.

Discussion

87068.	A solution of weak acid HA containing 0.01 moles of acid per litre of solutions has pH = 4. The percentage degree of ionisation of the acid and the ionisation constant of acid are respectively
Answer» `1%, 10^(-6)` `0.01%, 10^(-4)` `1%, 10^(-4)` `0.01%, 10^(-6)` SOLUTION :`H^(+) = C ALPHA` `alpha = (H^(+))/(C) = (10^(-4))/(10^(2)) = 10^(-6)`.

Discussion

87069.	A solution of urea (molecular mass = 60 g mol^(-1))boils at 100.18^(@)C at atmospheric pressure. If K_(f) "and" K_(b) for water are 1.86 and 0.512 K kg mol^(-1) respectively, the above solution will freeze at :
Answer» `0.654^(@)C` `-0.654^(@)C` `6.54^(@)C` `-6.54^(@)C` SOLUTION :`(DeltaT_(f))/(DeltaT_(B))=K_(f)/K_(b)` `DeltaT_(f)=DeltaT_(b)=(0.18^(@))xx(1.86)/(0.512)=0.654^(@)C` Freezing POINT of solution = `0-0.654^(@)C=-0.654^(@)C`

Discussion

87070.	A solution of urea (molar mass 56) boils at 100.18^@C at atmospheric pressure. If K_f and K_g for water are 1.86 and 0.512 K "molality"^(-1) respectively, the above solution will freeze at
Answer» `-6.54^@C` `6.54^@C` `-0.654^@C` `0.654^@C` ANSWER :C

Discussion

87071.	A solution of urea received from some research laboratory has been marked mole fraction (x) and molality (m) at 10^@C. While calculating its molality and mole fraction in the laboratory at 24^@C, you will find
Answer» Mole FRACTION (X) and MOLALITY (m) Mole fraction (2X) and molality (2M) Mole fraction (X/2) and molality (m/2) Mole fraction (X) and molality (2m) Answer :A

Discussion

87072.	A solution of urea (mol. mass 60g mol^(-1)) boils at 100.18^(@)C at the atmospheric pressure. If K_(f) and K_(b) for water are 1.86 and 0.512 kg mol^(-1) respectively the above solution will freeze at
Answer» `-6.54^(@)C` `6.54^(@)C` `0.654^(@)C` `-0.654^(@)C` Solution :`DELTA T_(b)=0.18 , Delta T_(b)=m K_(b)` `(0.18)/(Delta T_(f))=(mK_(b))/(m K_(f)) , (0.18xx1.86)/(0.512)=Delta T_(f) , Delta T_(f)=0.653` `T^(+)-T_(s)=0.653 , T_(s)=0-0.653^(@)C , T_(s)=-0.653^(@)C`

Discussion

87073.	A solution of urea (mol. Mass "56 g mol"^(-1)) boils at 100.18^(@)C at the atmospheric pressure. If K_(f) and K_(b) for water are "1.86 and 0.512 K kg mol"^(-1) respectively, the above solution will freeze at
Answer» `-6.54^(@)C` `-0.654^(@)C` `6.54^(@)C` `0.654^(@)C` Solution : `DeltaT_(b)=K_(b)xxm therefore0.18=0.512xxm` `or m = 0.18//0,512` `DeltaT_(F)=K_(f)xxm=1.86xx(0.18)/(0.512)=0.654` `therefore""T_(f)=-0.654^(@)C`

Discussion

87074.	A solution of urea (mol. mass = 56 g mol""^(-1)) boils at 100.78^@ C at the atmospheric pressure. If K_f and K_b for water are 1.86 and 0.512 K kg mol-respectively, then the above solution will freeze at:
Answer» `0.654^@ C` ` - 0.654^@ C` `6.54^@ C` `- 6.54^@ C` SOLUTION :` - 0.654^@ C`

Discussion

87075.	A solution of urea (mol. Mass 56 g mol ^(-1)) boils at 100.18^(@)C at the atmospheric pressure. If K_(f) and K_(b) for water are 1.86 and 0.512 K kg mol ^(-1) respectively, the above solution will freeze at
Answer» `0.654^(@)C` `-0.654^(@)C` `6.54^(@)C` `-6.54^(@)C` SOLUTION :As`Delta T_(f) =K_(f)` m `Delta T_(b) =K_(b).m` Hence, we have `m =(Delta T_(f))/(K _(f)) =(Delta T_(b))/(K_(b))` or `Delta T _(f) = Delta T_(b) = (K_(f))/(K _(b))` `implies [Delta T _(b) =100. 18-100= 0.18^(@)C]` `= 0.18xx (1.86)/(0.512) =0.654^(@)C` As the Frecezing POINT of pure WATER is `0^(@)C, Delta T _(f) =0-T_(f)` `0.654 =-0T_(f)` `therefore T _(f) =-0.654` Thus the FREEZING point of solution will be `- 0.654 ^(@)C.`

Discussion

87076.	A solution of urea ( mol. =56 g "mol"^(-1) ) boils at 100.18^(@)C at the atmospheric pressure. If K_(f) and K_(b) for water are 1.86 and 0.512 K kg "mol"^(-1) respectively, then the above solution will freeze at :
Answer» `0.654^(@)C` `-0.654^(@)C` `6.54^(@)C` `-6.54^(@)C` SOLUTION :`m=(DeltaT_(F))/(K_(f))=(Delta T_(b))/(K_(b))` `Delta T_(b)=100.18-100=0.18^(@)` `:.DeltaT_(f)=(Delta T_(b))/(K_(b))xxK_(f)` `=(0.18)/(0.512)xx1.86=0.654^(@)` `:.` FREEZING point of solution `=0-0.654` `=-0.654^(@)C`

Discussion

87077.	A solution of urea in water has a boiling point of 100.128^(@)C. Calculate the freezing point of the same solution. Molal constants for water K_(f) and K_(b) are 1.86^(@)C and 0.512^(@)C respectively.
Answer» Solution :Step I. To calculate MOLALITY of the solution from boiling point data : We are GIVEN that`""DeltaT_(b)=100.128-100=0.128^(@)C, K_(b)=0.512^(@)C` ltbRgt Using the formula,`""DeltaT_(b)=K_(b)xxm`, where m is the molality, we get `""m=(DeltaT_(b))/(K_(b))=(0.128)/(0.512)=0.25` Step II. To caluate the DEPRESSION in freezing point : We are given that `K_(F)=1.86^(@)C` m = 0.25 (calculated above) `therefore""DeltaT_(f)=K_(f)xxm=1.86xx0.25=0.465^(@)` `therefore"Freezing point of the solution "(T_(f))=T_(f)^(@)-DeltaT_(f)=0^(@)C-0.465^(@)=-0.65^(@)C`

Discussion

87078.	A solutionof urea in water has a boiling point 101.128^(@)C .Calculatethe freezingpoint of the samesolution. Molal constant for water ,k_(f) and k_(b) are 1.86K m^(-1) and0.512K m^(-1) respectively.
Answer» Solution :`DeltaT_(B) = 101.128 - 100 = 1.128^(@)` ` m = (DELTA T_(b))/(k_(b)) = (1.128)/(0.512) = 2.203 m` ` Delta T_(F) = k_(f) xxm = 1.86 xx 2.203 = 41` Freezingpoint of solution` = 0 - 4.1= - 4.1^(@)C`

Discussion

87079.	A solution of urea in water boills at 100^(@)C. Calculate the freezing point of the same solution. Molal constant K_(f) and K_(b) are 1.86 and 0.521 k m^(-1) repectivly.
Answer» Solution :`DeltaT_(f)=K_(f)xxm" and "DeltaT_(b)xxm` `(DeltaT_(f))/(DeltaT_(b))=K_(f)/K_(b)` `DeltaT_(b)=100.18^(@)C-100.0=0.18^(@)C=0.18 K` `K_(f)=1.86" K m"^(-1), K_(b)=0.512" K m"^(-1)` `DeltaT_(f)=K_(f)/K_(b)xxDeltaT_(b)=((1.86"Km"^(-1)))/((0.512"Km"^(-1)))xx0.18K=0.654K=0.654^(@)C` `therefore "Freezing point of soluion"=0^(@)C-0.654^(@)C=-0.654^(@)C.`

Discussion

87080.	A solution of urea contain 8.6 gm/litre (mol. wt. 60.0). If is isotonic with a 5% solution of a non-volatile solute. The molecular weight of the solute will be
Answer» `348.9` `34.89` 3489 `861.2` Solution :For two non-electrolytic solution if isotonic, `C_(1)=C_(2)` `therefore (8.6)/(60xx1)=(5xx1000)/(m.wt. xx 100) therefore m = 348.9`

Discussion

87081.	A solution of two liquids boils at a temprature more than the boiling point of either of them .Hence the binary solution show
Answer» Negative deviation from Raoult's LAW Positive deviation from Raoult's law No deviation from Raoult's law Positive or negative deviation UPON the composition Solution :SOLUTIONS BOIL at relatively higher temprature that means,vapour PRESSURE of solution is less then it shows negative deviation from Raoult.s law.

Discussion

87082.	A solution of the two liquids A and B obeys Raoult's law. At a certain temperature, it is found that when the total pressure above the given solution is 400 mmHg, the mole fraction of A is 0.45 and that in the liquid is 0.65, what are vapour pressures of the two liquids?
Answer» SOLUTION :277 MM, 629 mm

Discussion

87083.	A solution of the given mixture was prepared in conc. HCI. On diluting this solution with water, a turbidity appears. This indicates the presence of
Answer» `As^(3+)` `HG^(2+)` `Sb^(3+)` `CU^(2+)` Answer :C

Discussion

87084.	A solution of sulphuric acid in water exhibits:
Answer» A) The APPLICABILITY of Henry's law B) Positive deviation from RAOULT's law C) Ideal properties D) Negative deviation from Raoult's law Answer :D

Discussion

87085.	A solution of sucross (molar mass = 342 g "mol"^(-1)) has been prepared by dissolving 68.4 g of sucrose in one kg of water. K_(f) for water is "1.86 K kg mol"^(-1) and vapour pressure of water at 298 K is 0.024 atm. If on dissolving the above amount of NaCl in 1 kg of water, the freezing point is found to be -0.344^(@)C, the percentage dissociation of NaCl in the solution is
Answer» `75%` `80%` `85%` `90%` SOLUTION :`DeltaT_(f)"(calculated)"=K_(f)xxm=1.86xx(5.85)/(58.5)=0.186^(@)` `DeltaT_(f)"(OBSERVED)"=0.344^(@)C THEREFORE i=(0.344)/(0.186)=1.85` `underset(1-ALPHA)(NaCl)rarrunderset(alpha)(Na^(+))+underset(alpha)(Cl)` `=1+alpha or alpha=i-1=0.85=85%`

Discussion

87086.	A solution of sucross (molar mass = 342 g "mol"^(-1)) has been prepared by dissolving 68.4 g of sucrose in one kg of water. K_(f) for water is "1.86 K kg mol"^(-1) and vapour pressure of water at 298 K is 0.024 atm. The freezing point of the solution will be
Answer» `-0.684^(@)C` `-0.342^(@)C` `-0.372^(@)C` `-0.186^(@)C` Solution :`DeltaT_(f)=K_(f)xxm=1.86xx(68.4)/(342)=0.372^(@)` `THEREFORE"Freezing POINT of the solution "=-0.372^(@)C`

Discussion

87087.	A solution of sucross (molar mass = 342 g "mol"^(-1)) has been prepared by dissolving 68.4 g of sucrose in one kg of water. K_(f) for water is "1.86 K kg mol"^(-1) and vapour pressure of water at 298 K is 0.024 atm. The mass of sodium chloride that should be dissolved in the same amount of water to get the same freezing point will be
Answer» 136.8 g 32.2 g 5.85 g 11.60 g Solution :`DeltaT_(f)=iK_(f)m therefore0.372=2xx1.86xxm or m =0.1` THUS, 0.1 mole, i.e., 5.85 g of NACL should be dissolved in 1 KG of water.

Discussion

87088.	A solution of sucross (molar mass = 342 g "mol"^(-1)) has been prepared by dissolving 68.4 g of sucrose in one kg of water. K_(f) for water is "1.86 K kg mol"^(-1) and vapour pressure of water at 298 K is 0.024 atm. the osmotic pressure of the solution at 298 K will be
Answer» 4.29 ATM 4.49 atm 4.69 atm 4.89 atm Solution :`pi=CRT=(68.4)/(342)xx0.082xx298="4.887 atm"`

Discussion

87089.	A solution of sucross (molar mass = 342 g "mol"^(-1)) has been prepared by dissolving 68.4 g of sucrose in one kg of water. K_(f) for water is "1.86 K kg mol"^(-1) and vapour pressure of water at 298 K is 0.024 atm. The vapour pressure of the solution at 298 K will be
Answer» 0.230 ATM 0.233 atm 0.236 atm 0.0239 atm Solution :`(p^(@)-p_(s))/(p^(@))=(n_(2))/(n_(1))=(w_(2)//M_(2))/(w_(1)//M_(1))` i.e., `(0.024-p_(s))/(0.024)=(68.4//342)/(1000//18)` `or(1000)/(18)(0.024-p_(s))=(68.4)/(342)xx0.024` `or 0.024-p_(s)=0.000086 or p_(s)=0.0239 atm.`

Discussion

87090.	A solution of sucrose (molecular mass 342/mol) is prepared by dissolving 68.4 g of it per litre of solution. What is the osmotic pressure at 300 K ? (R = 0.082 lit. atm "deg"^(-1)"mol"^(-1)).
Answer» SOLUTION :4.92 ATM

Discussion

87091.	A solution of sucrose (molar mass = 342 g/mol) is prepared by dissolving 68.4 g of it per litre of the solution, what is its osmotic pressure (R =0.082 "lit. atm." k^(-1)mol^(-1))at 273 K
Answer» 6.02 ATM 4.48 atm 4.04 atm 5.32 atm Solution :`pi=CRT` `pi=(w XX R xx T)/(MV)=(68.4xx0.0821xx273)/(342)=4.48` atm

Discussion

87092.	A solution of sucrose (molar mass 342 "gmol"^(-1)) has been produced by dissolving 68.5g sucrose in 1000g water. The freezing point of the solution obtained will be (K_f for H_2O= 1.86 kg "mol"^(-1) )
Answer» `-0.372^@C` `-0.520^@C` `+0.372^@C` `-0.570^@C` ANSWER :C

Discussion

87093.	A solution of sucrose (molar mass =342 g mol^(-1)) has been prepared by dissolving 68.5 g of sucrose in 100 g of water. The freezing point of the solution obtained will be (K_(f) "for water =1.86 K kg " mol^(-1)):
Answer» `-0.372^(@)C` `-0.52^(@)C` `+0.372^(@)C` `-0.57^(@)C` Solution :MOLALITY (m) of solution `=("Mass of solute 1 KG of solvent")/"MOLAR mass"` `=((68.5g)//(1kg))/((342gmol^(-1)))=0.2 MOL//kg` `DeltaT_(f)=K_(f)xxm` ` =(1.86 K kg mol^(-1))xx(0.2 mol//kg^(-1))` f.p. of solution =`(0-0.372)=-372^(@)C`.

Discussion

87094.	A solution of sucrose (molar mass =324 g/mol) is prepared by dissolving 68.4 g in 1000 g of water. What is the freezing point and boiling point of the solution ? K_(f)= water is 1.86 K/mand K_(b) =0.52 K/m.
Answer» SOLUTION :`M_(B)=342" g mol"^(-1), W_(B)=68.4 g, W_(A)=1 KG,` `K_(f)=1.86Km^(-1), K_(b)=0.52Km^(-1),T_(f)=?, T_(b)=?` `DeltaT_(f)=(K_(f)xxW_(B))/(M_(B)xxW_(A))""DeltaT_(f)=(K_(b)xxW_(B))/(M_(B)xxW_(A))` `=((186" K kg mol"^(-1))xx(68.4 g))/((342" g mol"^(-1))xx(1kg)),""DeltaT_(b)=((0.52" K kg mol"^(-1))xx(68.4g))/((342g)xx(1 kg))` `=0.372 K,"" =0.104 K` `T_(f)=T_(f)^(@)-DeltaT_(f)=273-0.372,""T_(b)=T_(b)^(@)+DeltaT_(b)=373+0.104` `=272.628 K""=373.104 K`

Discussion

87095.	A solution of sodium sulphate in water is electrolyzed using inert electrodes. The products at the cathode and anode are respectively
Answer» `O_(2), SO_2` `O_(2), H_2` `O_(2) , NA` `H_(2), O_(2)` ANSWER :D

Discussion

87096.	A solution of sodium thiosulphate on addition of few drops of ferric chloride gives violet colour due to the formation of
Answer» `Na_(2)S_(4)O_(6)` `Na_(2)SO_(4)` `Fe_(2)(S_(2)O_(3))` `Fe_(2)(S_(2)O_(3))_(2)` Answer :C

Discussion

87097.	A solution of solute 's' in benzene boils at 0.126^(@) higher than benzene. The molality of the solution is ( K_(b) for benzene = 2.52Km^(-1)) :
Answer» 2 m 0.05 m 1 m 20 m Solution :`DeltaT_(B)=K_(b)XX m ` or `m=(DeltaT_(b))/(K_(b))=(0.126)/(2.52)=0.05`.

Discussion

87098.	A solution of sodium sulphate in water is electrolyzed using inert electrodes. The products at the cathode and anode are respectively:
Answer» `H_(2),O_(2)` `O_(2),H_(2)` `O_(2),NA` `O_(2),SO_(2)` ANSWER :A

Discussion

87099.	A solution of sodium sulphate in water is electrolysed using inert electrodes. The productsat the cathode and anode are respectvely:
Answer» `H_2,O_2` `O_2H_2` `O_2,NA` `O_2,SO_2` ANSWER :A

Discussion

87100.	A solution of sodium saltof fatty acid was electrolyzed during Kolbe's reaction . The solution leftafter electrolysis is
Answer» Richer in `H_2SO_4` Richerin sodium SALT Richerin NAOH All Answer :C

Discussion

Explore topic-wise InterviewSolutions in .

A solution when dilutedwith H_(2)O And bolled gives a white precipitate .On the additionof excess NH_(2)CINH_(4)OH the volumeof theprecipitatedecreases leavingg bebinda whitege3lationtiousprecipitateidentifytheprecipitatewhichdissolves in NH_(4)OH//NH_(4)Cl :

A solution when diluted with H_(2)O and boiled, it gives a white precipitate. On addition of excess NH_(4)Cl//NH_(4)OH, the volume of precipitate decreases leaving behind a white gelatinous precipitate. Identify the precipitate which dissolves in NH_(4)OH//NH_(4)Cl.

A solution when diluted with H_(2)O and boiled, it gives a white precipitate. On addition of excess NH_(4)Cl//NH_(4)OH the volume of precipitate decreases leaving behind a white gelatinous precipitate. Identify the precipitate which dissolves in NH_(4)OH//NH_(4)Cl

A solution that obeys Raoult's law is called

A solutiion that contain more solute than the saturated solution at a given termperature ________ solution .

A solution prepared by dissolving 8.95 mg of a gene fragment in35.0 ml of water has an osmotic pressure of 0.335 ton at 25^(@)C. Assuming thegene fragment is a non-electrolyse, determine the molar mass.

A solution prepared by dissolving 8.95 mg of a gene fragment in 35.0 mL of water has an osmotic pressure of 0.335 torr at 25^(@)C. Assuming that the gene fragment is a non-electrolyte, calculate its molar mass.

A solution prepared by dissolving 15 g of non-volatile solute in 270 g of water gave relative lowering of vapour pressure of 0.005. The molecular weight of the solute is

A solution prepared by dissolving 1.25 g of oil of winter green (methyl salicylate) in 99.0 g of benzene has a boiling point of 80.31^(@)C. Determine the molar mass of this compound (B.P. of pure benzene = 80.10^(@)C and K_(b) for benzene =2.53^(@)"C kg mol"^(-1).

A solution prepared by dissolving 1.25 g of oil of winter green (methyl s.ilicylate) in 99.0 g of benzene has a boiling po . int of 80.31^(@) C , Determine the molar mass of this compound. (B.P. of pure benzene = 80.10^(@)C and K_(b) for benzene=2.53^(@) C " Kg " mol^(-1) )

A solution prepared by dissolving 1.25g of oil of winter green (methyl sallicylate) in 99.0g of benzene has a boiling point of 80.31^(@)C. Determine the molar mass of this compound. (B.P. of pure benzene =80.10^(@)C and K_(b) for benzene =2.53^(@)C kg "mol".1)

A solution prepared by dissolving 0.8gm of naphthalene in 100g of C Cl_4 has a boiling point elevation of 0.4^@C. A 1.24 g of an un known solute in same amount of C Cl_4 produced boiling point elevation of 0.62^@C, then molar mass of un-known solute is

A solution of white crystals with a solubleof Na_(2)CO_(3) .The action of cone H_(2)SO_(4) on the crystals yieds a brown gas .The crystalsare of

A solution on treatment with NH_(3) turns blue, contains

A solution of white crystals gives a precipitate with AgNO_(3) but no precipitate with a solution of Na_(2)CO_(3). The action of conc. H_(2)SO_(4) on the crystals yields a brown gas. The crystals are of :

A solutionof weakbase BOHwas titrated with 0.1 N HCl . The pHof the solution was foundto be10.24and 9.14 after the additionof 5 ml and20ml of the acidrespectively. Findthe dissociation constantof the base .

A solution of weak acids is diluted by adding an equal volume of water. Which of the following will not change

A solution of weak acid HA containing 0.01 moles of acid per litre of solutions has pH = 4. The percentage degree of ionisation of the acid and the ionisation constant of acid are respectively

A solution of urea (molecular mass = 60 g mol^(-1))boils at 100.18^(@)C at atmospheric pressure. If K_(f) "and" K_(b) for water are 1.86 and 0.512 K kg mol^(-1) respectively, the above solution will freeze at :

A solution of urea (molar mass 56) boils at 100.18^@C at atmospheric pressure. If K_f and K_g for water are 1.86 and 0.512 K "molality"^(-1) respectively, the above solution will freeze at

A solution of urea received from some research laboratory has been marked mole fraction (x) and molality (m) at 10^@C. While calculating its molality and mole fraction in the laboratory at 24^@C, you will find

A solution of urea (mol. mass 60g mol^(-1)) boils at 100.18^(@)C at the atmospheric pressure. If K_(f) and K_(b) for water are 1.86 and 0.512 kg mol^(-1) respectively the above solution will freeze at

A solution of urea (mol. Mass "56 g mol"^(-1)) boils at 100.18^(@)C at the atmospheric pressure. If K_(f) and K_(b) for water are "1.86 and 0.512 K kg mol"^(-1) respectively, the above solution will freeze at

A solution of urea (mol. mass = 56 g mol""^(-1)) boils at 100.78^@ C at the atmospheric pressure. If K_f and K_b for water are 1.86 and 0.512 K kg mol-respectively, then the above solution will freeze at:

A solution of urea (mol. Mass 56 g mol ^(-1)) boils at 100.18^(@)C at the atmospheric pressure. If K_(f) and K_(b) for water are 1.86 and 0.512 K kg mol ^(-1) respectively, the above solution will freeze at

A solution of urea ( mol. =56 g "mol"^(-1) ) boils at 100.18^(@)C at the atmospheric pressure. If K_(f) and K_(b) for water are 1.86 and 0.512 K kg "mol"^(-1) respectively, then the above solution will freeze at :

A solution of urea in water has a boiling point of 100.128^(@)C. Calculate the freezing point of the same solution. Molal constants for water K_(f) and K_(b) are 1.86^(@)C and 0.512^(@)C respectively.

A solutionof urea in water has a boiling point 101.128^(@)C .Calculatethe freezingpoint of the samesolution. Molal constant for water ,k_(f) and k_(b) are 1.86K m^(-1) and0.512K m^(-1) respectively.

A solution of urea in water boills at 100^(@)C. Calculate the freezing point of the same solution. Molal constant K_(f) and K_(b) are 1.86 and 0.521 k m^(-1) repectivly.

A solution of urea contain 8.6 gm/litre (mol. wt. 60.0). If is isotonic with a 5% solution of a non-volatile solute. The molecular weight of the solute will be

A solution of two liquids boils at a temprature more than the boiling point of either of them .Hence the binary solution show

A solution of the two liquids A and B obeys Raoult's law. At a certain temperature, it is found that when the total pressure above the given solution is 400 mmHg, the mole fraction of A is 0.45 and that in the liquid is 0.65, what are vapour pressures of the two liquids?

A solution of the given mixture was prepared in conc. HCI. On diluting this solution with water, a turbidity appears. This indicates the presence of

A solution of sulphuric acid in water exhibits:

A solution of sucross (molar mass = 342 g "mol"^(-1)) has been prepared by dissolving 68.4 g of sucrose in one kg of water. K_(f) for water is "1.86 K kg mol"^(-1) and vapour pressure of water at 298 K is 0.024 atm. The freezing point of the solution will be

A solution of sucross (molar mass = 342 g "mol"^(-1)) has been prepared by dissolving 68.4 g of sucrose in one kg of water. K_(f) for water is "1.86 K kg mol"^(-1) and vapour pressure of water at 298 K is 0.024 atm. the osmotic pressure of the solution at 298 K will be

A solution of sucross (molar mass = 342 g "mol"^(-1)) has been prepared by dissolving 68.4 g of sucrose in one kg of water. K_(f) for water is "1.86 K kg mol"^(-1) and vapour pressure of water at 298 K is 0.024 atm. The vapour pressure of the solution at 298 K will be

A solution of sucrose (molecular mass 342/mol) is prepared by dissolving 68.4 g of it per litre of solution. What is the osmotic pressure at 300 K ? (R = 0.082 lit. atm "deg"^(-1)"mol"^(-1)).

A solution of sucrose (molar mass = 342 g/mol) is prepared by dissolving 68.4 g of it per litre of the solution, what is its osmotic pressure (R =0.082 "lit. atm." k^(-1)mol^(-1))at 273 K

A solution of sucrose (molar mass 342 "gmol"^(-1)) has been produced by dissolving 68.5g sucrose in 1000g water. The freezing point of the solution obtained will be (K_f for H_2O= 1.86 kg "mol"^(-1) )

A solution of sucrose (molar mass =342 g mol^(-1)) has been prepared by dissolving 68.5 g of sucrose in 100 g of water. The freezing point of the solution obtained will be (K_(f) "for water =1.86 K kg " mol^(-1)):

A solution of sucrose (molar mass =324 g/mol) is prepared by dissolving 68.4 g in 1000 g of water. What is the freezing point and boiling point of the solution ? K_(f)= water is 1.86 K/mand K_(b) =0.52 K/m.

A solution of sodium sulphate in water is electrolyzed using inert electrodes. The products at the cathode and anode are respectively

A solution of sodium thiosulphate on addition of few drops of ferric chloride gives violet colour due to the formation of

A solution of solute 's' in benzene boils at 0.126^(@) higher than benzene. The molality of the solution is ( K_(b) for benzene = 2.52Km^(-1)) :

A solution of sodium sulphate in water is electrolyzed using inert electrodes. The products at the cathode and anode are respectively:

A solution of sodium sulphate in water is electrolysed using inert electrodes. The productsat the cathode and anode are respectvely:

A solution of sodium saltof fatty acid was electrolyzed during Kolbe's reaction . The solution leftafter electrolysis is