Answer (4) :
Molar mass of Glucose = 180g/mol
Molar mass of CaCl2 = 111 g/mol
Both the solution are prepared by dissolving 0.275 mole of solute in 2500 g of solvent (water) .
w/w(%) = mass of solute / mass of solution *100
Mass of solute (glucose) taken = moles * molar mass
Mass of Glucose taken= 0.275 * 180 = 49.5 g
Mass of solution (glucose) = mass of solute + mass of solvent = 49.5 + 2500 = 2549.5 g
w/w (%) for glucose solution=( 49.5 /2549.5 )*100
= 1.94%
Mass of CaCl2 taken = moles * molar mass
Mass = 0.275 * 111 = 30.525 g
Mass of solution of CaCl2 = 30.525+2500 = 2530.525 g
w/w(%) = (30.525 / 2530.525)*100 = 1.2%
Molality of solution = moles of solute / mass of solvent (in kg)
Since both solution contain equal no. of moles in equal mass of solvent .So, molality of both solution is same
Molality of Glucose solution = Molality of CaCl2 solution
(2500 g of water means 2.5 Kg of water (solvent))
Molality of solution = 0.275 / 2.5 = 0.11 m
=> To calculate melting point and boiling point of each solution
As depression in freezing point is calculated as
∆Tf = i* Kf * m ( i = van't Hoff factor determined by no. of ions formed by dissociation of solute )
For glucose = i = 1 (no dissociation)
For CaCl2 => i = 3 ( 3 ions formed due to dissociation of CaCl2 in water .i.e 1 Ca2+ and 2 Cl- )
Kf = 1.86 °CKg/mol
So, ∆Tf = i * Kf *m and ∆Tf = T°f - Tf
T°f is melting of pure water
Tf is melting point of solution (formed by dissolving electrolyte)
Also in case of boiling point : ∆ Tb = i * Kb * m
Kb = 0.51 °C Kg/mol
Also ∆Tb = Tb - T°b ( T°b = boiling point of solution)
Tb = boiling point of pure solvent water
∆Tf = i * Kf * m | Tf = T°f - ∆Tf | ∆Tb = i * Kb * m | Tb = ∆Tb + T°b | |
Glucose solution | 1*1.86 * 0.11 = 0.2046 °C | Tf = 0 - 0.2046 = -0.2046 °C | 1* 0.51 * 0.11 = 0.0561 °C | Tb = 0.0561 + 100 = 100.0561 °C |
CaCl2 solution | 3* 1.86 * 0.11 = 0.6138 °C | Tf = 0-0.6138 = -0.6138 ° C | 3* 0.51 * 0.11 = 0.1683 °C | Tb = 0.1683 + 100 = 100.1683 °C |
From the above calculation :
Melting point of glucose solution = -0.2046 °C
melting point of CaCl2 solution= -0.6138 °C
Boiling point of glucose solution = 100.0561 °C
Boiling point of CaCl2 solution = 100.1683 °C
So, boiling point of CaCl2 is more and freezing point of glucose solution is more .
Answer (5): Molar mass of KMnO4 is 158 g/mol
Mass of solute taken = 0.415 g
Moles of KMnO4 = mass taken / molar mass
Moles = 0.415 / 158 mol
Volume of solution = 250 ml = 0.25 Litre
Molarity of solution= moles of solute / volume of solution (L)
Molarity =( 0.415/ 158 )/ 0.25 = 0.0105 M
So, Molarity of first solution = 0.0105 M
Second solution :
made by taking 25 ml of first solution and diluted to 100 ml
Dilution factor = volume of initial solution / total volume of new solution = 25/100 = 0.25
Molarity of second solution = Molarity of first solution* dilution factor
Molarity of second solution = 0.0105 * 0.25 = 0.002625 M
= 2.625 * 10-3 M
Third solution :
made by taking 10 ml of second solution and diluted to 50 ml
Dilution factor = 10/ 50 = 0.2
Molarity of third solution = Molarity of second solution * dilution factor
Molarity of third solution = 2.625 * 10-3 * 0.2 = 5.25 * 10-4 M
Need help with these both please Two aqueous soiutions are prepared in which one contains 0.275 moles of the nonvolatile so glucose (ClO dissolved in 2500.00 g of water and the other solution con...
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