Question

Calculate the molality of a solution formed by dissolving 34.8 g of lii in 500.0 ml of water. calculate the molality of a solution formed by dissolving 34.8 g of lii in 500.0 ml of water. 0.520 m 0.260 m 0.254 m 0.696 m 0.130 m

Answer 1

First step is to find the number of moles in 34.8 gm of LiI:
From the periodic table: molar mass of Li = 6.941 while that of I = 126.9
Molar mass of LiI = 6.941 + 126.9 = 133.841
Number of moles = mass/molar mass = 34.8/133.841 = 0.26 moles

Second step is to calculate the molality:
molality = number of moles of solute / kilogram of solvent
At STP: 1L of water = 1Kg, thus, 500 ml of water = 0.5 kg
molality = 0.26/0.5 = 0.52 mole/Kg

Answer 2

The molality of solution formed by dissolving 34.8 g of LiI in 500 mL of water is $\boxed{0.520{\text{ m}}}$.

Further Explanation:

Concentration terms are used to determine concentration of various components present in any mixture. Some of these are mentioned below.

1. Molarity (M)

2. Mole fraction (X)

3. Molality (m)

4. Parts per million (ppm)

5. Mass percent ((w/w) %)

6. Volume percent ((v/v) %)

7. Parts per billion (ppb)

Molality is defined as moles of solute that are present per kilograms of solvent. It is represented by m and its unit is mol/kg. The expression for molality of solution is as follows:

${\text{Molarity of solution}} = \dfrac{{{\text{Moles }}\left( {{\text{mol}}} \right){\text{of solute}}}}{{{\text{Mass }}\left( {{\text{kg}}} \right){\text{ of solvent}}}}$  

Given information:

Mass of LiI: 34.8 g

Volume of water: 500.0 mL

To calculate:

Molality of LiI solution

How to proceed:

Step1: Moles of LiI that are present in 34.8 g of LiI has to be determined. This is done with the help of equation (1).

The formula to calculate the moles of LiI is as follows:

${\text{Moles of LiI}} = \dfrac{{{\text{Mass of LiI}}}}{{{\text{Molar mass of LiI}}}}$                                            …… (1)

The mass of LiI is 34.8 g.

The molar mass of LiI is 133.841 g/mol.

Substitute these values in equation (1).

$\begin{aligned}{\text{Moles of LiI}} &= \frac{{{\text{34}}{\text{.8 g}}}}{{{\text{133}}{\text{.841 g/mol}}}} \\&= 0.26{\text{ mol}} \\\end{aligned}$  

Step 2: Mass of water is to be evaluated.

At standard temperature and pressure conditions, one liter of water can be considered equivalent to 1 kilograms of water.

Therefore mass of water can be calculated as follows:

$\begin{aligned}{\text{Mass of water}} &= \left( {500.0{\text{ mL}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ L}}}}{{1{\text{ mL}}}}} \right)\left( {\frac{{1{\text{ kg}}}}{{1{\text{ L}}}}} \right) \\&= 0.5{\text{ kg}} \\\end{aligned}$  

Step 3: Molality of LiI solution is to be calculated. This is done with the help of equation (2).

The formula to calculate molality of LiI solution is as follows:

${\text{Molality of LiI solution}} = \dfrac{{{\text{Moles of LiI}}}}{{{\text{Mass of water}}}}$                                                …… (2)

Substitute 0.26mol for moles of LiI and 0.5 kg for mass of water in equation (2).

$\begin{aligned}{\text{Molality of LiI solution}} &= \frac{{{\text{0}}{\text{.26 mol}}}}{{{\text{0}}{\text{.5 kg}}}} \\&= 0.520{\text{ m}} \\\end{aligned}$  

Hence, molality of given solution comes out to be 0.520 m.

Learn more:

1. Calculation of volume of gas: brainly.com/question/3636135
2. Determine how many moles of water produce: brainly.com/question/1405182

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Concentration terms

Keywords: concentration, concentration terms, solutions, molarity, molality, LiI, moles, mass, molar mass, 0.520 m, 0.5 kg, 0.26 mol.