Question

A solution of ethanol, ${{C}_{2}}{{H}_{6}}O$, is prepared by dissolving 14.0g ${{C}_{2}}{{H}_{6}}O$ in 100.0 g of water. (Molar mass ethanol = 46.7g/mole)
Find the molality of the solution.
Find the % (m/m) concentration of the solution.

Answer 1

Hint: We know the mathematical formula for Molality and Mass percentage. Use these formulas to calculate. Here ethanol is the solute and water is the solvent.
1. Molality
2. Mass percentage(m/m) %

Complete step by step solution:
We will first define what molality is, in order to get a clear idea.
Molality – Molality is defined as the number of moles of solute dissolved in 1kg of solvent.
A solution with a molality of 1mol/kg is also defined as 1 molal or 1m.
It is independent of temperature.
Its unit is – mole/kg
Given mass of ${{C}_{2}}{{H}_{6}}O$ is 14g
Molar mass of ${{C}_{2}}{{H}_{6}}O$ is given that is 46.07/mole
$Molality=\dfrac{Number\,of\,moles\,of\,solute}{Mass\,of\,solvent\,in\,kg}$.
$Molality$ = $\dfrac{14\times 1000}{46.07\times 100}$ = $\dfrac{140}{46.07}$
= 3.03 m $\approx$ 3 m
So, the calculated molality of the solution is 3 mol/Kg.
Here also we will first define what Mass percentage is, in order to get a clear idea.
Mass percentage (m/m) % - Mass of solute dissolved in 100 g of solution is called mass percentage.
Mass of the solute is given as 14g
Mass of the solvent given as 100g
Mass of the solution = Mass of solute + Mass of solvent
= 14g + 100g = 114g

}\%=\dfrac{Mass\,of\,solute}{Mass\,of\,solution}\times 100$$ $$Mass\text{ }percentage\left( m/m \right)\text{ }\%=\dfrac{14}{114}\times 100=12.28\%$$ So, we have got the % (m/m) concentration of the solution is 12.28. Note: Don’t get confused between molality and molarity, we should always keep in mind that molarity is defined as the number of moles of solute dissolved in 1 litre of solution. And make sure that our units are in the SI system.