Henry's law constant for CO 2 in water is 1 . 67 × 10 8 Pa at 298 K . Calculate the quantity of CO 2 in 500 ml of soda water when packed under 2 . 5 atm CO 2 pressure at 298 K .

Given, Henry's law constant ( K H ) = 1 . 67 × 10 8 Pa Partial pressure of dissolved gas ( P CO 2 ) = 2 . 5 atm = 2 . 5 × 101325 Pa x co 2 = P co 2 K H = 2 . 5 × 101325 1 . 67 × 10 8 = 1 . 517 × 10 - 3 Number moles of water = 500 18 = 27 . 78 moles n H 2 O = 27 . 78 moles X CO 2 = n CO 2 n CO 2 + n H 2 O = n CO 2 n H 2 O = n CO 2 27 . 78 n CO 2 = X CO 2 × 27 . 78 = 0 . 0422 mol

If nitrogen gas is passed in water at 293 K , then how many milli mole N 2 gas will be soluble in one litre water? The partial pressure of N 2 is 0 . 987 bar and the value of Henry's law constant K H at 293 K is 76 . 48 bar .

According to Henry's law the partial pressure and mole fraction of dissolved N 2 gas are given by x N 2 = p N 2 K H = 0 . 987 76 . 48 × 10 3 bar = 1 . 29 × 10 - 5 In a very dilute solution, mole fraction of solute B is given by x B = n B n B + n A = n B n A Mass of 1 litre water = w A = 1000 mL × 1 g mL - 1 = 1000 g Molar mass of water = M A = 18 g mol - 1 Mole of water = n A = w A M A = 1000 g 18 g mol - 1 = 55 . 56 mol On substituting the values of x B and n A into equation (i), we get n B = x B × n A = 1 . 29 × 10 - 5 × 55 . 56 / mol = 7 . 1672 × 10 - 4 mol = 0 . 7167 m mol

On dissolving 4 gm urea in 100 gm water, 1 . 24 ℃ depression in freezing point of the solution is obtained. Calculate molecular mass of urea. ( K f = 1 . 86 K mol - 1 )

The molal freezing point depression constant ( K f ) of solvent is defined as the lowering of freezing point caused by 1 mole of a solute dissolved in 1 kg of the solvent. Given, K f = 1 . 86 K mol - 1 The relation between molar mass of solute, M B and depression in freezing point, ∆ T f is given by the equation, M B = K f ∆ T f × w B w A Where, w B = mass of solute, w A =mass of solvent Hence, molecular mass of urea, M B = 1 . 86 × 4 1 . 24 × 100 × 1000 M B = 60 g mol - 1

In a sample of drinking water, many impurities are found including chloroform, CHCl 3 , which are carcinogenic. The level of these impurities was 15 ppm (according to mass) Find the molality of chloroform in this sample of water.

Molalility is defined as the number of moles of solute in 1000 g of solvent. The molality of the solution does not change with the change in temperature of the solution. 15 ppm of chlorine means that 15 g of chlorine e is present in 10 6 g of solution. Molality of chloroform is = Mass of chloroform Molar mass of chloroform × ( Mass of solution - mass of chloroform ) × 1000 = 15 119 . 5 × ( 10 6 - 15 ) × 10 3 = 1 . 255 × 10 - 4 m .

In a sample of drinking water, many impurities are found including chloroform, CHCl 3 , which are carcinogenic. The level of these impurities was 15 ppm (according to mass) Show it in mass percentage

When a solute is present in a very small amount in a solution, its concentration is expressed in terms of parts per million. 15 ppm corresponds to 15 g of impurities in 10 6 g of solution. We have, Mass percentage = Mass of impurities Mass of solution × 100 Mass percentage = 15 10 6 × 100 = 1 . 5 × 10 - 3 %

Write the definition of the following : Molality

Molality can be defined as the number of moles of solute dissolved in 1 kg of a solvent. Mathematically, this can be expressed as, molality = Mass of solute Molecular mass of solute × mass of solvent in kg .

Write the definition of the following : Molarity

The number of moles of solute present in one litre of solution is called molarity. It is represented by M . Molarity = Number of moles of solute Volume of solution in Litre = n V Molarity = weight of the solute Gram molecular weight of solute × 1 Volume of solution ( litres ) Molarity can be used to convert between the moles of solute or mass and the volume of the solution.

Write the definition of the following : Mole fraction

Mole fraction:- It is defined as the ratio of the number of moles of the particular component in the solution to the total number of moles of all the components present in the solution. If A is the number of moles of solute dissolved in B moles of solvent, then Mole fraction of solute X A = moles of solute Total moles of solution = n A n A + n B X B = moles of solvent Total moles of solution = n B n A + n B X A + X B = n A n A + n B + n B n A + n B = n A + n B n A + n B = 1 X A + X B = 1 or X A = 1 - X B or X B = 1 - X A

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter -1 - Solutions>Madhya Pradesh Board-2020>Q 19

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Find molality of pure water $(H_{2} O = 18)$

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