Question

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 contains 0.275 moles of the nonvolatile solute CaCI, dissolved in 2500.00 g of water. Describe how each of the colligative properties listed below would be affected for each of these solutions (for each olatile solut 4. is a decrease or an increase then identify which of the solutions (CafinOs or CaCh) would cause the greater effect). In addition, calculate the changes and resulting value for either the melting point or boiling point for each of these solutions (choose onhz I either the melting point or boiling point for each solution then follow worked example 9.15 and 9.16 in your text). Show how to determine (and then calculate) the concentration of each solution in both % (w/w) and molality. melting point of each solution (the normal melting point of H:O is 0.000 C& k-18600C.kg/mol) boiling point of each solution (the normal boiling point of H:0 is 100.000 Ca K-05100 -kg/mol) s. An aqueous solution is prepared by dissolving 0.415 g of KMnO, in 250.0 mL of water. This solution is then diluted by taking a 25.0 mL aliquot (sample) and diluting it to 100.0 ml, with water. A third solution is then prepa 0 mL with water. Show how to determine (and then calculate) the molarity of each solution (the 3 different solutions prepared) The molarity of the first solution (initial solution) is: The molarity of the second solution (1M dilution) is: The molarity of the third solution (2rd dilution) is:

Answer 1

Answer (4) :

Molar mass of Glucose = 180g/mol

Molar mass of CaCl_​​​2 = 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 CaCl_​​​2 taken = moles * molar mass

Mass = 0.275 * 111 = 30.525 g

Mass of solution of CaCl_​​​2 = 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 CaCl_​​​2 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

∆T_f = i* K_{​​​​​​f} * m ( i = van't Hoff factor determined by no. of ions formed by dissociation of solute )

For glucose = i = 1 (no dissociation)

For CaCl_​​​2 => i = 3 ( 3 ions formed due to dissociation of CaCl_​​​2 in water .i.e 1 Ca^{​​​​​2+} and 2 Cl^{​​​​​-} )

K_{​​​​​​f} = 1.86 °CKg/mol

So, ∆T_f = i * K_{​​​​​​f} *m and ∆T_f = T°_f - T_{​​​​​​f}   

T°_f is melting of pure water

T_{​​​​​​f} is melting point of solution (formed by dissolving electrolyte)

Also in case of boiling point : ∆ T_{​​​​​​b} = i * K_{​​​​​​b} * m  

K_{​​​​​​b} = 0.51 °C Kg/mol

Also ∆T_b = T_{​​​​​​b} - T°_b  ( T°_b = boiling point of solution)

T_{​​​​​​b =} boiling point of pure solvent water

	∆Tf = i * K​​​​​​f * m	T​​​​​​​​​​f = T°f - ∆Tf ​​	∆Tb = i * K​​​​​​b * m	T​​​​​​b = ∆Tb + T°b
Glucose solution	1*1.86 * 0.11 = 0.2046 °C	T​​​​​​f = 0 - 0.2046 = -0.2046 °C	1* 0.51 * 0.11 = 0.0561 °C	T​​​​​​b = 0.0561 + 100 = 100.0561 °C
CaCl​​​2 solution	3* 1.86 * 0.11 = 0.6138 °C	T​​​​​​f = 0-0.6138 = -0.6138 ° C	3* 0.51 * 0.11 = 0.1683 °C	T​​​​​​b = 0.1683 + 100 = 100.1683 °C

From the above calculation :

Melting point of glucose solution = -0.2046 °C

melting point of CaCl_​​​2 solution= -0.6138 °C

Boiling point of glucose solution = 100.0561 °C

Boiling point of CaCl_​​​2 solution = 100.1683 °C

So, boiling point of CaCl_​​​2 is more and freezing point of glucose solution is more .

Answer (5): Molar mass of KMnO_​​​4 is 158 g/mol

Mass of solute taken = 0.415 g

Moles of KMnO_​​​4 = 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