Solutions Test 1

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Solutions Test 1

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Question 1
1 / -0

Molality is expressed in

A
mol/kg

B
mol/L

C
L/mol

D
g/L

Solution

Molality = moles of solute / kg of solvent.
Question 2
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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?

A
1000 g

B
10⁶ g

C
100 g

D
1.0 g

Solution

1 ppm is equivalent to 1 part out of 1 million (10⁶) 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
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Mole fraction of glycerine, C₃H₅(OH)₃ in a solution containing 36 gm of water and 46 gm of glycerine is:

A
0.40

B
0.20

C
0.46

D
0.36

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

A
2.85m

B
3.15m

C
1.45m

D
2.50m

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
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When the solvent is in solid state, solution is

A
Solid solution

B
Gaseous solution

C
Solution

D
Liquid solution

Solution

When both the solute and solvent are in solid state, then the solution is called as solid solution or solid sol.
Question 6
1 / -0

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

A
27 moles

B
1 moles

C
9 moles

D
3 moles

Solution

In 3M there will be 3 moles per litre therefore in 3 litres there will be 9 moles.
Question 7
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A sample of 300.0 g of drinking water is found to contain 38 mg Pb. What this concentration in parts per million?

A
3 x 10² ppm

B
6.5 m

C
130 ppm Pb

D
21 ppm

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
10⁶= 1,000,000
parts of solution. You can thus say that a 1 ppm solution will contain exactly 1 g of solute for every 10⁶g of solution.
In your case, you know that you have

in exactly
300.0 g = 3.000 ⋅ 10²g solution

This means that you can use this known composition as a conversion factor to scale up the mass of the solution to 10⁶g

Since this represents the mass of lead present in exactly 10⁶g of solution, you can say that the solution has a concentration of
Question 8
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What is the mass of NaCl required to prepare 0.5 liters of a 2.5 molar solution of NaCl?

A
37.80 g

B
73 g

C
117 .00g

D
58.50 g

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

A
1.25 M

B
2.5 M

C
1.00 M

D
0.875 M

Solution

The molarity of a resulting solution is given by
Question 10
1 / -0

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.

A
0.03, Ethyl chloride

B
0.08, cannot be determined

C
0.03, Methanol

D
0.08, Ethyl chloride

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.