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251.

Two aliphatic aldehydes P and Q react in th presence of aqueous `k_(2)CO_(3)` to give compound R, which upon treatment with HCN provides compound S. On acidification and heatingm,S gives the product shown below: The compound R isA. B. C. D.

Answer» Correct Answer - A
Discussion
252.

Two aliphatic aldehydes P and Q react in th presence of aqueous `k_(2)CO_(3)` to give compound R, which upon treatment with HCN provides compound S. On acidification and heatingm,S gives the product shown below: The compound S isA. B. C. D.

Answer» Correct Answer - D
Discussion
253.

Under the same reaction conditions, the intial concentration of `1.386 mol dm^(-3)` of a substance becomes half in `40 s` and `20 s` theough first order and zero order kinetics, respectively. The ratio `(k_(1)//k_(0))` of the rate constants for first order `(k_(1))` and zero order `(k_(0))` of the reaction isA. `0.5 mol^(-1) dm^(3)`B. `1.0 mol dm^(-3)`C. `1.5 mol dm^(-3)`D. `2.0 mol^(-1) dm^(3)`

Answer» Correct Answer - A
Discussion
254.

Four solid sphereas each of diameter `sqrt(5) cm` and mass `0.5 kg` are placed with their centres at the corners of a square of side `4 cm`. The moment of inertia of the system about the diagonal of the square is `N xx 10^(-4) kg-m^(2)`, the `N` is -

Answer» Correct Answer - 9
Discussion
255.

Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes the freezing point of the solution. Use the freezing point depression constant of water as 2 K `kg mol^(-1).` The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T). [molecular weight of ethanol is` 46 g mol^(-1)` Among the following, the option representing change in the freezing point isA. B. C. D.

Answer» Correct Answer - C
Discussion
256.

The standard state Gibbs free energies of formation of ) C(graphite and C(diamond) at T = 298 K are `Delta_(f)G^(@)["C(graphite")]=0kJ mol^(-1)` `Delta_(f)G^(@)["C(diamond")]=2.9kJ mol^(-1)` The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite [ ) C(graphite ] to diamond [C(diamond)] reduces its volume by `2xx10^(-6)m^(3) mol^(-1).` If ) C(graphite is converted to C(diamond) isothermally at T = 298 K, the pressure at which ) C(graphite is in equilibrium with C(diamond), is `["Useful information:"1J=1kg m^(2)s^(-2),1Pa=1kgm^(-1)s^(-2),1"bar"=10^(5)Pa]`A. 14501 barB. 58001 barC. 1450 barD. 29001 bar

Answer» Correct Answer - A
Discussion
257.

For the following cell, `Zn(s)|ZnSO_(4)(aq)||CuSO_(4)(aq)||Cu(s)` When the concentration of `Zn^(2+)` is 10 times the concentration of `Cu^(2+)`, the expression for `DeltaG` (in J `"mol"^(-1)`) [F is Faraday constant, R is gas constant] T is temperaure, `E^(@)("cell")=1.1V`A. `1.1F`B. `2.300RT-2.2F`C. `2.303RT+1.1F`D. `-2.2F`

Answer» Correct Answer - B
Discussion
258.

Monoatomic gas A having 5 mole is mixed with diatomic gas B having 1 mole in container of volume `V_(0)`. Now the volume of mixture is compressed to `(V_(0))/(4)` by adiabatic process. Initial pressure and temperature of ags mixture is `P_(0)` and `T_(0)`. [given `2^(3.2)=9.2`] Choose correct option :A. `._(gamma mix)=1.6`B. Final pressure is between `9P_(0)` and `10P_(0)`C. `|W.D|=13RT_(0)`D. none of these

Answer» `._(gamma mix)=(n_(1)C_(P1)+n_(2)C_(P2))/(n_(1)C_(V1)+n_(2)C_(V2))=(8)/(5)`
`W = (P_(1)V_(1)-P_(2)V_(2))/(gamma-1)`
`P_(0)V_(0)^(8//5)=P_(2)((V_(0))/(4))^(8//5)`
`P_(2)=9.2P_(0)`
`W = (P_(0)V_(0)-9.2P_(0)(V_(0))/(4))/(3//5)= - 13RT_(0)`
Discussion