In 214 . 2 g syrup of sugar, 34 . 2 g sugar ( C 12 H 22 O 11 ) is present. What will be the molality of sugar syrup? ( C = 12 , H = 1 , O = 16 ).

Given, The weight of sugar syrup is 214 . 2 g . The weight of sugar is 34 . 2 g . The weight of water, i.e., solvent is 214 . 2 - 34 . 2 = 180 g . The molality will be m = number of moles of solute Weight of solvent ( 1 kg ) . The number of moles of sugar will be 34 . 2 12 × 12 + 22 × 1 + 11 × 16 = 34 . 2 342 = 0 . 1 mole. The molality is 0 . 1 180 × 1000 = 100 180 = 0 . 56 .

A decrease of 1 . 2 ° C in freezing point took place on dissolving 7 . 5 g of substance in 75 g of water. Find out the molecular weight of the substance. (Molal depression constant for water k ' f = 1 . 86 ° C ).

Given, Depression in freezing point = 1 . 2 ° C . Mass of solute = 7 . 5 g . Mass of water = 75 g = 0 . 075 kg ( 1 kg = 1000 g ) . Formula used : ∆ T f = K f × m . ∆ T f = i × K f × Mass of solute Molar mass of solute × Mass of water in Kg . Where, ∆ T f is change in freezing point. K f is freezing point constant. ' m ' is molality. Now put all the given values in this formula, we get, 1 . 2 = 1 . 86 × 7 . 5 g M × 0 . 075 g . M = 1 . 86 × 7 . 5 1 . 2 × 0 . 075 = 155 g mol - 1 . The molecular weight of substance is 155 g mol - 1 .

The boiling point of pure acetone is 56 . 38 ° C . When 0 . 707 gm of an organic compound is dissolved in 10 g of acetone, there is elevation to 56 . 88 ° C in boiling point. What is the molecular weight of the compound? Molal elevation constant ( k b ) for acetone is 16 . 7 ° .

Given, The boiling point of pure acetone, = 56 . 38 ° C . The elevation in boiling point of acetone = 56 . 88 ° C . Weight of the solute, = 0 . 707 g . Molal elevation constant, k b = 16 . 7 ° . The relationship between molal elevation constant and the molality is given as, ΔT b = K b × m . ( 56 . 88 − 56 . 38 ) = 1 . 67 × 0 . 707 M B × 1000 10 . M B × 0 . 50 = 1 . 67 × 0 . 707 × 100 . M B = 236 . 13 g mol - 1 . The molecular mass of the given organic compound is 236 . 13 g mol - 1 .

A solution of glucose containing 0 . 45 gram in 250 mL water was prepared at 0 ° C . What is its osmotic pressure? R = 0 . 0821 L . atm . / degree / mol .

Osmotic pressure is proportional to the molarity, C , of the solution at a given temperature T . Thus, π = CRT . Where, π = osmotic pressure. R = gas constant = 0 . 0821 dm 3 atm mol - 1 K - 1 . T = temperature = 298 K . C = molarity = n V s = W B M B × 1000 250 . Where, n = no. of moles. W B = Weight of solute glucose = 0 . 45 g . M B = Molecular weight of solute glucose = 180 g . V = Volume of solution = 250 mL . C = W B M B × 1000 250 = 0 . 45 180 × 1000 250 = 0 . 01 . The osmotic pressure, π = CRT = 0 . 01 × 0 . 0821 × 298 = 0 . 244 atm . The osmotic pressure is 0 . 244 atm .

Calculate the molality of 20 % (by weight) potassium carbonate solution. K = 39 , C = 12 , O = 16 .

Given, Weight of potassium carbonate, K 2 CO 3 = 20 % which means 20 g of K 2 CO 3 in 100 g of solution. So, weight of solvent = ( weight of solution ) - ( weight of solute ) = 100 - 20 = 80 g = 0 . 08 kg . We know that, the molality will be, molality = ( moles of solute ) ( mass of solvent in kg ) . molality = 20 2 × 39 + 12 + 3 × 16 × 1 0 . 08 . molality = 20 2 × 39 + 12 + 3 × 16 × 1 0 . 08 = 20 11 . 04 = 1 . 81 . The molality is 1 . 81 .

What are isotonic solutions? Explain with example.

Two solutions having the same osmotic pressure across a semipermeable membrane is referred to as isotonic solution. It has the same osmolarity (solute concentration), as another solution. A solution is isotonic when its effective mole concentration is the same as that of another solution. This state provides the free movement of water across the membrane without changing the concentration of solutes on either side. Some examples of isotonic solutions are 0 . 9 % normal saline and lactated ringers. When a solution has a higher solute concentration than the solution present across the semi-permeable membrane, it is known as a hypertonic solution.

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter -1 - Solutions>Uttar Pradesh Board-2020>Q 8

EASY

12th Uttar Pradesh Board

IMPORTANT

Earn 100

In which of the following lowering of vapour pressure would be maximum?

50% studentsanswered this correctly

Chapter-Wise 3D Chemistry Videos from Embibe Experts EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY

07 min 28 sec

Elevation in Boiling Point

The boiling point of a liquid is the temperature at which its vapour pressure becomes equal to atmospheric pressure. Let’s learn about the elevation in boili...

05 min 12 sec

Degree of Association

The video shows the property of aggregation of solute particles when dissolved in some particular solvents. Learn more about the degree of the association th...

04 min 21 sec

Degree of Dissociation

The video introduces the concept of degree of dissociation and its importance in the study of colligative properties. Calculating the degree of dissociation...

Learn and Prepare for any exam you want !

|

|

|

|

|

|

|

|