Answer
Verified
389.1k+ views
Hint: The molarity of the solution can be calculated by dividing the number of moles with the volume of the solution. The molality of the solution is calculated by dividing the number of moles of the solute with the mass of the solvent in kg.
Complete step by step answer:
Molarity: The molarity of the solution is defined as the number of moles of the solute present per liter. It is represented by the symbol, M.
\[Molarity=\frac{\text{Moles of the solute}}{\text{Volume of the solution}}\]
Moles of the solute can be calculated by:
\[Moles=\frac{\text{mass of the solute}}{\text{molar mass of the solute}}\]
Molality: The molality of a solution is defined as the number of moles of the solute dissolved in 1kg (1000 g) of the solvent. It is represented by the symbol, ‘m’.
\[Molality=\frac{\text{Moles of the solute}}{\text{Mass of the solvent in kg}}\]
Now, according to the question: the mole fraction of $C{{H}_{3}}OH$ in an aqueous solution is 0.02.
So, $0.02=\frac{x}{y+x}$
Where x = moles of $C{{H}_{3}}OH$ and y = moles of water.
Moles of water = $\frac{1000\text{ g}}{18}=55.55$
Hence,
$0.02=\frac{x}{55.55+x}$
$x-0.02x=55.55\text{ x 0}\text{.02}$
$\text{0}\text{.98}x=1.111$
$x=1.13\text{ moles}$
So, the moles of solute ($C{{H}_{3}}OH$ ) is 1.13 and mass of solvent is 1kg.
Therefore, the molality of the solution is $\frac{1.13}{1}=1.13\text{ m}$
Now, for molarity of the solution,
The moles of the solute ($C{{H}_{3}}OH$ ) is 1.13.
The volume of the solution can be calculated by dividing the total mass of the solution with the density. Given density = $0.994g/c{{m}^{3}}$
Total mass = molecular mass of $C{{H}_{3}}OH$ x moles of$C{{H}_{3}}OH$ + molecular mass of water x moles of water
Total mass = $32\text{ x 1}\text{.13 + 18 x 55}\text{.55 = 1036}\text{.06}$
Volume = $\frac{1036.06}{0.994}mL=\frac{1036.06}{994}L=1.04L$
Now, putting these in molarity equation,
Molarity = $\frac{1.13}{1.04}=1.08M$
Therefore, the molarity of the solution is 1.08 M and the molality of the solution is 1.13 m.
Note: The volume of the solution must be taken in liters only. The value of a mole fraction of water can directly be taken as 55.55 because the aqueous solution will have water as the solvent.
Complete step by step answer:
Molarity: The molarity of the solution is defined as the number of moles of the solute present per liter. It is represented by the symbol, M.
\[Molarity=\frac{\text{Moles of the solute}}{\text{Volume of the solution}}\]
Moles of the solute can be calculated by:
\[Moles=\frac{\text{mass of the solute}}{\text{molar mass of the solute}}\]
Molality: The molality of a solution is defined as the number of moles of the solute dissolved in 1kg (1000 g) of the solvent. It is represented by the symbol, ‘m’.
\[Molality=\frac{\text{Moles of the solute}}{\text{Mass of the solvent in kg}}\]
Now, according to the question: the mole fraction of $C{{H}_{3}}OH$ in an aqueous solution is 0.02.
So, $0.02=\frac{x}{y+x}$
Where x = moles of $C{{H}_{3}}OH$ and y = moles of water.
Moles of water = $\frac{1000\text{ g}}{18}=55.55$
Hence,
$0.02=\frac{x}{55.55+x}$
$x-0.02x=55.55\text{ x 0}\text{.02}$
$\text{0}\text{.98}x=1.111$
$x=1.13\text{ moles}$
So, the moles of solute ($C{{H}_{3}}OH$ ) is 1.13 and mass of solvent is 1kg.
Therefore, the molality of the solution is $\frac{1.13}{1}=1.13\text{ m}$
Now, for molarity of the solution,
The moles of the solute ($C{{H}_{3}}OH$ ) is 1.13.
The volume of the solution can be calculated by dividing the total mass of the solution with the density. Given density = $0.994g/c{{m}^{3}}$
Total mass = molecular mass of $C{{H}_{3}}OH$ x moles of$C{{H}_{3}}OH$ + molecular mass of water x moles of water
Total mass = $32\text{ x 1}\text{.13 + 18 x 55}\text{.55 = 1036}\text{.06}$
Volume = $\frac{1036.06}{0.994}mL=\frac{1036.06}{994}L=1.04L$
Now, putting these in molarity equation,
Molarity = $\frac{1.13}{1.04}=1.08M$
Therefore, the molarity of the solution is 1.08 M and the molality of the solution is 1.13 m.
Note: The volume of the solution must be taken in liters only. The value of a mole fraction of water can directly be taken as 55.55 because the aqueous solution will have water as the solvent.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Select the correct plural noun from the given singular class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
The sum of three consecutive multiples of 11 is 363 class 7 maths CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How many squares are there in a chess board A 1296 class 11 maths CBSE