Relation Between Molarity and Molality: Definition & Formula

Ans. Molarity is a measure of the concentration of a solution. It is defined as the number of moles of solute present in one liter of the solution. Molarity is expressed in units of moles per liter (mol/L).

Ques. Explain the steps to determine the molality of the solution using the given molarity. (3 marks)

Ans. To determine the molality (m) of a solution with known molarity (M), the following steps are performed:

1. Determine the density (ρ): Measure the mass and volume of a known quantity of the solution. Calculate the density using the formula density = mass/volume.
2. Use the relation formula: Substitute the known values of molarity (M), density (ρ), and molar mass of solute (M′) into the relation formula to find the molality (m).

m = 1000M/{1000ρ – MM’}

These steps allow you to determine the molality of the solution based on the given molarity and experimentally determined density.

Ans. (c) 6

Explanation: Given that

• Weight of sodium ions given = 138 gm
• Molecular mass of sodium = 23 gm
• Number of moles of sodium ion = 138/23 = 6 moles
• Weight of solvent given = 1 kg

Hence,

Molality = number of moles of sodium ion/ weight of solvent in kg

= 6/1

= 6 m 

Ans. Density = Mass/volume

Volume = Mass/density

= 500/0.936 = 534.18 ml of solution

Molarity = number of moles of solute/ volume of solution in l

 \(=\frac{50}{58.44}* \frac {1000}{534.18}\)

=1.60 M

Ans. Molarity is the number of moles of solute present in 1 liter of the solution. Molality is the number of moles of solute in 1 kg of solvent. This clearly indicates molarity is a function of the volume of the solution.

• A change in the temperature of the system can hence affect the molarity of the solution.
• Molality is a function of the mass of the solution and thus has no effect on a change in temperature.
• This independence of molality on temperature makes it a more preferred choice for expressing solution concentrations. 

Ans. 1M means 1 mole of solute in 1L of solution. Here, we may note that the solution is a mixture of both solute and solvent. 

1m means 1 mole of solute in 1kg of solution. Here, the solution is 1L of solvent.

This implies that the quantity of solvent is more in 1m of the solution when compared to 1M of the solution (for the same 1 mole of solute). Thus the concentration of 1M solutions is greater than 1m of solution with the same number of moles of the solute. 

Ans. Normality is expressed as the gram equivalent of solute per liter of solution. It is another way of calculating the concentration of a solution like the molarity and molality. The S.I. unit is eq/L and is denoted by N. It can be calculated by the formula:

N = Weight of Solute in gm. / Equivalent Weight × Volume in ltr.

Ans. Mole fraction of water in solution = 0.85-mole fraction H₂​SO_4​ in solution =1−0.85=0.15.

If n₁​ is the number of moles of water and n₂​ is the number of moles of H_2​SO₄​ in the solution, then:

Mole fraction of H₂​SO₄​ = \(\frac{n_2}{n_1 + n_2}\) = 0.15

The molality of the H₂​SO₄​ solution means the number of moles of H₂​SO₄​ present in 1000 of H₂​O.

Therefore, we have, w₁​=1000

g or n₁​ = 1000/18 ​= 55.55, n₂​=1

n₂/55.55+n₂​ ​​= 0.15

n₂​ = 0.15n₂ ​+ 8.3325

or Molality (n₂) ​= 9.8 m

Ans. The volume of the solution is 1000 ml.

∴ Moles of NaCl=3 moles

∴ Mass of NaCl=3×(23+35.5)g

=175.5 g=0.1755 kg

∴ Density = Mass of solution​/Volume of solution

∴ Mass of solution = 1.26×1000

=1260 g = 1.26 kg

∴ Mass of solvent = Mass of solution- Mass of NaCl

=1.26−0.1755

=1.0845 kg

∴ Molality = Moles of NaCl​/Mass of solvent (kg)

=1.08453​

∴ Molality = 2.766 m

Ans. Let us assume that we have 100 g of solution.

Then the mass of C₂H₆O₂ in this solution will be 20 g (20% of 100 g). Similarly, the mass of water in this solution will be 80 g.

Molar mass of C₂H₆O₂

=2x12 + 6x1 + 2x16 = 24+6+32 = 62 g mol^-1

Molar mass of H₂O = 2x1+16 = 18 g mol^-1

Moles of ethylene glycol = (20g)/(62g\mol^(-1)) = 0.322 mol

Moles of water = (80g)/(18g\mol^(-1)) = 4.44 mol

Now, mole fraction can be calculated as follows: X_(glycol)=(n_g)/(n_g+n_w)

=(0.322)/(0.322+4.444) = 0.068

Molarity (M): The number of moles of solute dissolved in one liter of solution is called the molarity of the solute.

Ans. Molar mass of C₂H₄O₂ =12 xx 2 + 1 xx 4 + 16 xx 2 = 60 g mol^–1

Moles of C₂H₄O₂ = (2.5g)/(60g\mol^(-1))=0.0417 mol

Mass of benzene in kg = (75 g)/(1000 g\ kg^(–1)) = 75 xx 10^(–3) kg

Molality of C₂H₄O₂ =(0.0417 mol)/(75x10^(-3)kg)=0.556 mol kg^-1

Ans. 1 L of solution = 1000 mL = 1000 cm³

1.329 g/cm³ times 1000 cm³ = 1329 g (the mass of the entire solution)

1329 g minus 571.4 g = 757.6 g = 0.7576 kg (the mass of water in the solution)

571.4 g / 98.0768 g/mol = 5.826 mol of H₂SO₄

5.826 mol / 0.7576 kg = 7.690 m

Ans. 1) Molality = moles solute divided by kilograms solute:

0.449 mol/kg = x / 2.92 kg

x = 1.31108 mol of KBr

2) Moles times molar mass equals grams:

(1.31108 mol) (119.0023 g/mol) = 156 g KBr

3) Adding them up:

156 g KBr + 2920 g water = 3076 g total

Ans. A mole fraction of 0.100 for NaCl means the mole fraction of water is 0.900.

Let us assume a solution is presently made up of 0.100 moles of NaCl and 0.900 moles of water.

• Mass of water present \(\rightarrow \) (0.900 mol) (18.015 g/mol) = 16.2135 g
• Molality of solution \(\rightarrow \) 0.100 mol / 0.0162135 kg = 6.1677 m

Ans. Molality = moles of naphthalene/kilograms of benzene

(16.5 g / 128.1732 g/mol) / 0.0543 kg = 2.37 m

Ans. Let’s assume that we have 1.00L of this solution. Hydrogen peroxide has a molecular formula of H₂O₂.

Use the given density to find the mass of the solution.

1.00L⋅1000mL/1L⋅1.01g/mL=1010g

Next, find the mass of the hydrogen peroxide present in the solution.

1010g(0.152)=153.52g

Convert the mass of hydrogen peroxide into moles of hydrogen peroxide.

153.52g of H₂O₂⋅1 mole H₂O₂/31.04g H₂O₂=4.95 moles of H₂O₂

Molarity=moles of solute/liters of solution

Since we have 1.00L of the solution, the molarity is 4.95M.

Ans. 35mL of 5.0MCuSO₄ solution is needed to prepare 0.350L of 0.500MCuSO₄.  

We can use the formula M₁V₁ = M₂V₂

Thus, (5.00M₁)(V₁)=(0.350M₂)(0.500L)

V₁=0.035L=35mL

Ans. M_initial \(\times\) V_initial = M_final \(\times\) V_final

• M_initial = 7M
• V_initial = 65ml
• M_final = 2M
• V_final =?

Solve for V_final since we are looking for the final volume of water needed to dilute the existing solution:

V_final = M_initial \(\times\) V_initial/M_final

=7M \(\times\) 65ml/2M = 227.5ml