0.1M NaCl and0.1M glucose solution means that 0.1 moles of solute dissolve in 1L of solvent.
weight of solvent in both solution is same .
Dpression in freezing point can br calculated as :
Kf = molal depression constant
wB = weight of solute
wA= weight of solvent
MB = molar mass of the solute
mass of NaCl = 58.69g/mol
mass of glucose = 180g/mol
The depression in freezingpoint is inversly proprotional to molar mass of solute thus more is molar mass of solute lesser is the depression in freezing point .
Thus it can explain that the NaCl wil be more than the glucose.
Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1.
Molality (m) is defined as the number of moles of the solute per kilogram (kg) of the solvent and is expressed as:
Molarity (M) is defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution.
(a) Mol. mass of
Volume of solution = 4.3 L
(b) Number of moles present in 1000 ml of 0.5M H2SO4= 0.5 mol
therefore number of moles present in 30ml of 0.5M H2SO4=mol =0.015mol
therefore molarity =0.015/0.5L
thus molarity is 0.03M
Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
Let the total mass of the solution be 100g and mass of benzene be 30 g
therefore mass of tetrachloride= (100-30)g = 70g
Molar mass of benzene,