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# Molarity Questions

One of the most common units used to quantify the concentration of a solution is molarity (M), which shows the number of moles of solute per litre of solution (moles/Litre). Molarity is a term of measurement that can be used to determine the volume of a solvent or the amount of a solute.

The formula c1V1 = c2V2, where c is concentration and V is volume, can be used to describe the relationship between two solutions containing the same amount of moles of solute.

 Definition: Molarity, also known as molar concentration, is defined as the number of moles of solute present in a given number of litres of solution, or moles per litre.

## Molarity Chemistry Questions with Solutions

Q1: Differentiate between Solute and Solvent.

A solute is a component that dissolves in a solvent to generate a solution. Solutes can exist in three states: liquid, gaseous, and solid. Solutes usually make up a smaller proportion of a solution than solvents.

A solvent is a material that has the ability to dissolve other substances. Solvents can exist in three states: liquid, gaseous, and solid. Solvents are most typically employed as liquids.

Q2: Define the Mole fraction.

Mole fraction is defined as the number of moles of one component of a solution divided by the total number of moles of all of the components in the solution.

The sum of all the mole fractions of a solution’s components is always 1.

Q3: Define Mass Percentage.

The mass percentage of an element of a solution is defined as the mass of the solute in grammes present in 100 g of the solution. It’s written like this:

$$\begin{array}{l}Mass % of a component = \frac{Mass of component in solution}{Total mass of solution} x 100\end{array}$$

Q4: Two solutions of a substance (non-electrolyte) are mixed in the following manner. 480 mL of 1.5 M first solution + 520 mL of 1.2 M second solution. What is the molarity of the final mixture?

a) 1.20 M

b) 1.50 M

c) 1.344 M

d) 2.70 M

The correct option is ‘c’.

Explanation: The molarity of a mixture, Mmix, can be calculated using the following formula:

Mmix = (M1V1 + M2V2) / (V1 + V2) = (1.5 x 480) + (1.2 x 520)/(480 + 520) = (720 + 624)/1000 = 1.344 M

Q5: With increase in temperature, which of these changes?

a) molality

b) fraction of solute present in water

c) mole fraction

d) weight fraction of solute

d) The weight fraction of solute present in water will vary with temperature.

Explanation: The volume of a solution is affected by temperature. Thus, the formula for concentration that includes volume will change as temperature changes.

Q6: The density of a solution prepared by dissolving 120 g of urea (mol.mass = 60 u) in 1000 g of water is 1.15 g/mL. The molarity of this solution is:

a) 2.05 M

b) 1.02 M

c) 0.50 M

d) 1.78 M

The correct option is ‘a’.

Explanation: Molarity of a solution (M) = n/V

Where:

n = number of moles of urea = 120/60 = 2 mol

V = volume of the solution = mass of solution / density of solution = (120 + 1000) g / 1.15 g mL-1 = 974 mL = 0.974 L

Therefore:

M = 2 mol / 0.974 L = 2.05 mol.L-1

Note: Volume of solution should be expressed in litres.

Q7: The density (in g.mL–1) of a 3.60 M sulphuric acid solution that is 29% H2SO4 (Molar mass = 98 g.mol–1) by mass will be:

a) 1.45

b) 1.64

c) 1.88

d) 1.22

The correct option is ‘d’.

Explanation: The relation between Molarity, M and mass percent (%) is given by:

M = (% x 10 x d) / MW

Where:

MW = molecular weight of solute

d = density of solution

Therefore:

d = M x MW / (% x 10) = 3.60 x 98 / 29 x 10 = 1.216 g.mL-1

Q8: What is the difference between Molarity and Molality?

The difference between a solution and a solvent is a fundamental distinction between molality and molarity.

The ratio of the moles of a solute to the total litres of a solution is known as molarity. Both the solute and the solvent are present in the solution.

Molality, on the contrary, is the ratio of a solute’s moles to a solvent’s kilograms.

Molarity (M) Molality (m)
Measure of Concentration Concentration
Definition The moles of a solute per litre of a solution. The moles of a solute per kilogram of a solvent.
Units M m
Equation M = moles solute/litres solution m = moles solute/kg solvent
Ratio of moles to Volume (in litres) Mass (in kilograms)

Q9: What is the number of moles of solute in 250 mL of a 0.4 M solution?

There are 0.1 moles of solute present in a 250 mL of 0.4 M solution.

Explanation: To begin, understand that the molar concentration indicates the number of moles of the solute present in one litre of solution. In a 0.4 M solution, each litre of solution contains 0.4 moles of solute. Using dimensional analysis, we can figure out how many moles of solute are in 250 mL of solution.

250mL x (1L/1000mL) x (0.4mol/1L) = 0.1 mol.

Q10: Calculate the molarity of each of the following solutions:

1. 30 g of Co (NO3)2.6H2O in 4.3L of solution
2. 30 mL of 0.5M H2SO4 diluted to 500 mL.

Molarity = Moles of Solute / Volume of Solution in Litre

1. Molar mass of Co (NO3)2.6H2O (solute) = 59 + 2 (14 + 3 × 16) + 6 × 18 = 291 g.mol-1

∴ Moles of Co(NO3)2.6H2O = 30 / 291 mol = 0.103 mol

Therefore, molarity = 0.103 mol / 4.3 L = 0.023 M

1. Number of moles present in 1000 mL (Solvent) of 0.5 M H2SO4 (Solute) = 0.5 mol

∴ Number of moles present in 30 mL of 0.5 M H2SO4 = (0.5 x 30 ) / 1000 mol = 0.015mol

Therefore, molarity = 0.015 mol / 0.5 L = 0.03 M

Q11: How many moles of Na2CO3 are there in 10.0L of 2.0M solution?

Molarity is represented by the following equation:

Molarity = Moles of Solute / Volume of Solution in Litre

We already have the molarity and volume of solution, both of which have good units.

Let’s rewrite the equation to figure out how many moles there are. We may do this by multiplying both sides of the equation by the L solution. On the right side, the L solution cancels out, leaving the number of moles equal to the molarity times volume:

Moles of solute = Molarity x Volume of Solution in Litre = (10.0 L) (2.0M) = 20 moles.

Thus, there are 20 moles of Na2CO3 in 10.0L of 2.0M solution.

Q12: How many grams of hydrogen chloride (HCl) are required to prepare 4 Litre of 5M HCl in water?

Given:

Molarity = 5M

Volume = 4L

∴ Moles of HCl = Molarity x Volume = 5 x 4 = 20 moles.

1 mole of HCl = 36.5gm

∴ 20 moles of HCl = 36.5 x 20gm = 730 gm

Therefore, 730gm of HCl is required to prepare 4 Litre of 5M HCl in water.

Q13: How many grams of sodium dichromate should be added to a 50ml volumetric flask to prepare 0.025 molar sodium dichromate when the flask is filled to mark with water?

50 ml volumetric flask is filled upto the mark so, 50 ml is the final volume.

So, volume required = 50 ml

Concentration = 0.025 M

Molecular weight of sodium dichromate (Na2Cr2O7) = 261.97 g

So, for 1 M solution 261.97 g of Na2Cr2O7 in 1000 ml

∴ 0.025 M solution ‘X’ of Na2Cr2O7 in 1000 ml

∴ ‘X’ = (0.025 x 261.97)/1000 = 6.54 g

∴ 0.025 M solution 6.54 g of Na2Cr2O7 in 1000 ml

So, 0.025 M ‘X’ g of Na2Cr2O7 in 50 ml

= (50 x 6.54)/1000 = 0.3274 g of Na2Cr2O7

Thus, 0.3274 g of Na2Cr2O7 is required to prepare 0.025 M 50 ml sodium dichromate solution.

Q14: A solution is prepared by bubbling 1.56 grams of hydrochloric acid in water. Here, the volume of the solution is 26.8 mL. Calculate the molarity of the solution.

The chemical formula of hydrochloric acid = HCl

The chemical formula for Water = H2O

The molecular weight of HCl = 35.5 ×1 + 1×1 = 36.5 moles/gram

The molecular weight of H2O = 1 ×2 + 16 ×1 = 18 moles/gram

Given, mass of hydrochloric acid in the solution = 1.56 g

The number of moles of hydrochloric acid = $$\begin{array}{l}n_{T} = \frac{1.56}{36.5}\end{array}$$

nT = 4.27 x 10-2 mole

Now, given volume of the solution = 26.8 mL

Expressing the volume in terms of litres,

$$\begin{array}{l}Volume = \frac{26.8}{1000}\end{array}$$

Volume = 2.68 x 10-2 Litres

Now, calculating the molarity of the solution using the formula given above.

$$\begin{array}{l}Molarity = \frac{No. of moles of element}{Volume of solution in litres}\end{array}$$
$$\begin{array}{l}Molarity =\frac{4.27 x 10^{-2}}{2.68 x 10^{-2}}\end{array}$$

= 1.59 M

Thus, the molarity of the solution is 1.59 M.

Q15: Find the molarity of a solution with 0.2074g of CA(OH)2 in 40ml of solution.

Given, Mass of Ca(OH)₂ (solute) = 0.2074g

Volume of solution = 40mL

Molar mass of Calcium (Ca) = 40g

Molar mass of Oxygen (O) = 16g

Molar mass of Hydrogen (H) = 1g

∴ Molar mass of Ca(OH)2 = 40 + 16×2 + 1×2 = 40+32+2 = 74g

Now, number of moles in 0.2074g of Ca(OH)2 :

= Given Mass/Molar mass

$$\begin{array}{l}= \frac{0.2074}{74} = \frac{2074}{740000}moles\end{array}$$

Converting the volume of the solution from mL to L, we get,

=> 1L = 1000mL

=> 40mL = 40/1000

=> 4/100 L

Therefore,

Molarity of a solution = Moles of solute/Litres of solution

$$\begin{array}{l}=\frac{2074/740000}{4/100}\end{array}$$
$$\begin{array}{l}=\frac{2074}{740000} x \frac{100}{4}\end{array}$$
$$\begin{array}{l}=\frac{518.5}{7400}\end{array}$$

=0.07 M [approximately]

Thus, the molarity of the solution is 0.07 M.

## Practise Questions on Molarity

Q1: What is the molarity of a solution with a mass of solute 10 kg mass and 100 litre volume?

a) 0.1 molar

b) 1 molar

c) 10 molar

d) 100 molar

Q2: Which one is more concentrated? 1M or 1m of the solution? Give a reason.

Q3: The molarity of an aqueous solution of CaCl is defined as the

a) moles of CaCl per millilitre of solution

b) grams of CaCl per litre of water

c) grams of CaCl per millilitre of solution

d) moles of CaCl per litre of solution

Q4: If 2 moles of salt is dissolved to form 1 litre of solution, calculate the molarity of the solution.

a) 1 M solution

b) 1.5 M solution

c) 2 M solution

d) 2.5 M solution

Q5: To calculate the Molarity of a solution when the solute is given in grams and the volume of the solution is given in millilitres, you must first:

a) Convert grams to moles, but leave the volume of solution in millilitres

b) Convert volume of solution in millilitres to litres, but leave grams to moles

c) Convert grams to moles, and convert volume of solution in millilitres to litres

d) None of the above

Click the PDF to check the answers for Practice Questions.

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