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QUESTION: 1

Molality is expressed in

Solution:

Molality = moles of solute / kg of solvent.

QUESTION: 2

The level of contamination of chloroform was found to be 15 ppm. It means 15 g of chloroform is present in how many grams of solution?

Solution:

1 ppm is equivalent to 1 part out of 1 million (10^{6}) parts.

∴ Mass percent of 15 ppm chloroform in water

⇒ 1.5 x 10^{-3} g chloroform present in 100 g water

Thus, 15g chloroform will be present in water

QUESTION: 3

Mole fraction of glycerine, C_{3}H_{5}(OH)_{3} in a solution containing 36 gm of water and 46 gm of glycerine is:

Solution:

Molecular mass of glycerine = 92

Molecular mass of water = 18

**► **No. of moles of glycerine = 46/92 = 0.5 moles

**► **No. of moles of water = 36/18 = 2 moles

Thus, **Mole fraction of glycerine = No. of moles of glycerine/(No. of moles of glycerine + No. of moles of water)** = 0.5/(2 + 0.5) = 0.20

QUESTION: 4

Calculate the molality of 12.5% w/w sulphuric acid?

Solution:

12.5% w/w means 12.5 g in 100 g of solution.

**► **Weight of solvent = 100 g – 12.5 g = 87.5 g.

**► **Number of moles of sulphuric acid = 12.5/ 98 = 0.127 mol

⇒ molality = 0.127 x 1000/87.5 = 1.45 m

QUESTION: 5

When the solvent is in solid state, solution is

Solution:

When both the solute and solvent are in solid state, then the solution is called as solid solution or **solid sol**.

QUESTION: 6

The number of moles of KCl in 3 L of 3 M solution is

Solution:

In 3M there will be 3 moles per litre therefore in 3 litres there will be 9 moles.

QUESTION: 7

A sample of 300.0 g of drinking water is found to contain 38 mg Pb. What this concentration in parts per million?

Solution:

We use **parts per million** to express the concentrations of solutions that contain very, very small amounts, often called **trace amounts**, of a given solute.

More specifically, a solution's concentration in *parts per millions* tells you the number of parts of solute present **for every**

10^{6 }= 1,000,000

**parts of solution**. You can thus say that a 1 ppm solution will contain exactly 1 g of solute **for every** 10^{6}g of solution.

In your case, you know that you have

in exactly

300.0 g = 3.000 ⋅ 10^{2}g solution

This means that you can use this known composition as a conversion factor to *scale up* the mass of the solution to 10^{6}g

Since this represents the mass of lead present in exactly 10^{6}g of solution, you can say that the solution has a concentration of

QUESTION: 8

What is the mass of NaCl required to prepare 0.5 liters of a 2.5 molar solution of NaCl?

Solution:

The molecular weight of NaCl is 58.44 g/mol

So, one mole of NaCl weighs 58.44 g.

A 2.5 M solution is 2.5 moles per liter (Molarity is just the number of moles per liter).

Therefore, 0.5 L would contain 1.25 mol. Hence, you would need 1.25 × 58.44 g = 73 g.

QUESTION: 9

The molarity of a solution obtained by mixing 750 ml of 0.5(M) HCl with 250ml of 2(M) HCl will be?

Solution:

The molarity of a resulting solution is given by

QUESTION: 10

Mole fraction of ethyl chloride and methanol in a ternary solution is 0.6 and 0.32 respectively. What is the mole fraction of third component. Also identify the solvent in this ternary solution.

Solution:

Mole fraction of third component is 1 - (0.6 + 0.32) = 0.08.

Since ethyl chloride has** highest mole fraction** so it is the solvent.

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