Chapter 16.4, Problem 44LC

Interpretation Introduction

Interpretation: To evaluate the grams of Sodium bromide that must be dissolved in 400 grams of water to produce a 0.500 molal solution.

Concept introduction: Molality is defined as the measurement of the Number of moles of solute present in 1kg of solvent.

The formula to calculate the molality is as follows:

  $\text{Molality = }\frac{\text{Number of moles of solute}}{\text{(Kg) of solvent}}$

Answer to Problem 44LC

The required gram value of Sodium bromide is $20.6$ gram.

Explanation of Solution

Molality is defined as the measurement of the Number of moles of solute present in 1kg of solvent.

  $m=\frac{n(\text{moles})}{M(\text{Kg})}$

Where,

• $n$ represents the moles of solvent.
• $M$ represents the Mass of the solution.

Applying the rule of three, where the solution is $0.5$ molal.

In $1kg$ solution the number of moles $=0.5$ (moles of solute).

In $0.4kg$ number of moles will be $=\frac{0.4\times 0.5}{1}=0.2$

The molar mass of Sodium $=23g/mol$

The molar mass of Bromide $=80g/mol$

The molar mass of Sodium bromide $=23g/mol+80g/mol=103g/mol$

Applying the rule of three;

1 mole of sodium bromide contain $103$ grams

So 0.2 moles will contain the mass $=\frac{0.2\times 103}{1}$

  $\text{Mass}=20.6\text{ grams}$

Conclusion

The required gram value of Sodium bromide is $20.6$ gram.