2g of solute is dissolved in two different solvent A and B having 230g mass each Given that `K_b`(A):`K_b`(B)= 1:2 calculate the ratio of Delta`T_b`(A):delta`T_b`(B)

1g of a non-volatile non-electrolyte solute is dissolved in 100g of two different solvents A and B whose ebullioscopic constants are in the ratio of 1 : 5. The ratio of the elevation in their boiling points, (DeltaT_(b)(A))/(DeltaT_(b)(B)) is

1g of same non volatile solute is added to 100g of two different solvents A and B . K_(b) of A:B=1:5 Find out ((DeltaT_(b))_(A))/((DeltaT_(b))_(B))

If solution containing 0.15 g of solute dissolved in 15 g of solvent boils at a temperature higher by 0.216^(@)C than that of pure solvent, the molecular mass of the substance is (K_(b)=2.16^(@)C)

If t is parameter,A=(a sec t,b tan t) and B=(-a tan t,b sec t),O=(0,0) then be locus of the centroid of Delta OAB is

If t is parameter,A=(a sec t,b tan t) and B=(-a tan t,b sec t),O=(0,0) then be locus of the centroid of Delta OAB is

The boiling point T'_(b) of a solvent becomes T_(b) on addition of X_(1) mole fraction of solute. Heat of vaporisation of solvent is DeltaH_("vap") . The relation between elevation in b.pt. DeltaT_(b)=(T_(b)-T'_(b)) can be given by: (DeltaT_(b))/(T_(b)xxT'_(b))=-(2lnX_(1))/(DeltaH_(vap)) A graph plotted between log_(10)X_(1) vs (1)/(T_(b)) gives:

In a thermally isolated container, material A of mass m is placed against material B, also of mass m but at higher temperature. When thermal equilibrium is reached, the temperature changes Delta T_A and Delta T_B of A and B are recorded. Then the experiment is repeated, using A with other materials. All of the same mass m. The results are given in the table. Rank the four materials according to their specific heats, greatest first. {:("Experiment", "Temperature","Changes"),(1., DeltaT_A=+50 C^@,DeltaT_B=-50 C^@),(2.,Delta T_A=+10 C^@,Delta T_C=-20 C^@),(3.,Delta T_A=+2 C^@,Delta T_D=-40 C^@):}