A solution containing 1 . 23 g of calcium nitrate in 10 g of water, boils at 100 . 975 ° C at 760 mm of Hg . Calculate the Van’t Hoff factor for the salt at this concentration. ( K b for water = 0 . 52 K kg mol - 1 , mol. wt. of calcium nitrate = 164 g mol - 1 ).

Given, Observed boiling point is 100 . 975 ° C for the solution of calcium nitrate in water. Amount of calcium nitrate (solute) = 1 . 23 g . Quantity of water (solvent) = 10 g . K b for water = 0 . 52 K Kg mol - 1 and M.Wt. of Calcium nitrate = 164 g mol - 1 . No. of moles of solute = 1 . 23 164 = 0 . 0075 moles. So, molality, m = 0 . 0075 × 1000 10 = 0 . 75 m . ΔT b = K b × m = 0 . 52 K Kg mol - 1 × 0 . 75 m . ΔT b = 0 . 39 K . Now, observed boiling point = 100 . 975 ° C + 273 = 373 . 975 K . As, elevation in boiling point is 0 . 39 K , the calculated boiling point becomes = 373 . 975 K + 0 . 39 K = 374 . 365 K . Hence, van.t Hoff factor, i = Observed boiling point Calculated boiling point = 373 . 975 K 374 . 365 K = 0 . 999 .

An aqueous solution of a non-volatile solute freezes at 272 . 4 K , while pure water freezes at 273 . 0 K . Determine the following: The lowering of vapour pressure of water at 298 K . (Given: K f = 1 . 86 K kg mol - 1 , K b = 0 . 512 K kg mol - 1 and vapour pressure of water at 298 K = 23 . 756 mm of Hg ).

The number of solvent molecules escaping into vapour phase gets reduced and as a result the pressure exerted by the vapour phase is also reduced. Given, Vapour pressure of water is 23 . 756 mm of Hg . The lowering of vapour pressure is P 0 - P s P 0 = 760 - 23 . 756 760 = 736 . 244 760 = 0 . 97 .

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12th ICSE

IMPORTANT

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The freezing point of a solution containing $5.85 g$ of $NaCl$ in $100 g$ of water is $- 3 \cdot 348^{o} C$ . Calculate Van’t Hoff factor ‘ $i$ ’ for this solution. What will be the experimental molecular weight of $NaCl$ ? ( $K_{f}$ for water $= 1 \cdot 86 K kg {mol}^{- 1}$ )

Important Questions on Solutions

EASY

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 2 - Solutions>ICSE-2018>Q 4

An aqueous solution containing

12.48 g

of barium chloride in

1000 g

of water boils at

100.0832 ℃

. Calculate the degree of dissociation of barium chloride.
(

K_{b}

for water

= 0.52 K Kg {mol}^{- 1}

)

HARD

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 2 - Solutions>ICSE-2019>Q 6

Calculate the boiling point of urea solution when

6 g

of urea is dissolved in

200 g

of water. (

K_{b}

for water

= 0.52 K kg {mol}^{- 1}

, boiling point of pure water

= 373 K

, molecular weight of urea

= 60

MEDIUM

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 2 - Solutions>ICSE-2019>Q 7

When $0.4 g$ of acetic acid is dissolved in $40 g$ of benzene, the freezing point of the solution is lowered by $0.45 K$ . Calculate the degree of association of acetic acid. Acetic acid forms dimer when dissolved in benzene. ( $K_{f}$ for benzene $= 5.12 K kg {mol}^{- 1}$ , atomic mass: $C = 12, H = 1, O = 16$ ).

MEDIUM

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 2 - Solutions>ICSE-2019>Q 8

A solution is prepared by dissolving $9.25 g$ of non-volatile solute in $450 mL$ of water. It has an osmotic pressure of $350 mm$ of $Hg$ at $27 ° C$ . Assuming the solute is non-electrolyte, determine its molecular mass. ( $R = 0.0821 L atm K^{- 1} {mol}^{- 1}$ ).

MEDIUM

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 2 - Solutions>ICSE-2020>Q 9

The elevation in boiling point when $0.30 g$ of acetic acid is dissolved in $100 g$ of benzene is $0.0633 ° C$ . Calculate the molecular weight of acetic acid from this data. What conclusion can you draw about the molecular state of the solute in the solution? (Given $K_{b}$ for benzene $= 2.53 K kg {mol}^{- 1}$ , at. wt. of $C = 12, H = 1, O = 16$ ).

EASY

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 2 - Solutions>ICSE-2020>Q 10

Determine the osmotic pressure of a solution prepared by dissolving $0.025 g$ of $K_{2} {SO}_{4}$ in $2$ litres of water at $25 ° C$ , assuming that $K_{2} {SO}_{4}$ is completely dissociated.

( $R = 0.0821 L atm K^{- 1} {mol}^{- 1}; Mol . wt . of K_{2} {SO}_{4} = 174 g {mol}^{- 1}$ ).

EASY

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 2 - Solutions>ICSE-2020>Q 11

An aqueous solution of a non-volatile solute freezes at $272.4 K$ , while pure water freezes at $273.0 K$ . Determine the following: The molality of solution. (Given: $K_{f} = 1.86 K kg {mol}^{- 1}, K_{b} = 0.512 K kg {mol}^{- 1}$ and vapour pressure of water at $298 K = 23.756 mm of Hg$ ).

EASY

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 2 - Solutions>ICSE-2020>Q 12

An aqueous solution of a non-volatile solute freezes at $272.4 K$ , while pure water freezes at $273.0 K$ . Determine the following: Boiling point of solution. (Given: $K_{f} = 1.86 K kg {mol}^{- 1}, K_{b} = 0.512 K kg {mol}^{- 1}$ and vapour pressure of water at $298 K = 23.756 mm of Hg$ ).

The freezing point of a solution containing 5.85 g of NaCl in 100 g of water is −3·348oC. Calculate Van’t Hoff factor ‘i’ for this solution. What will be the experimental molecular weight of NaCl? (Kf for water = 1·86 K kg mol-1)

Important Questions on Solutions

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The freezing point of a solution containing $5.85 g$ of $NaCl$ in $100 g$ of water is $- 3 \cdot 348^{o} C$ . Calculate Van’t Hoff factor ‘ $i$ ’ for this solution. What will be the experimental molecular weight of $NaCl$ ? ( $K_{f}$ for water $= 1 \cdot 86 K kg {mol}^{- 1}$ )