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.
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,
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.