Question

Given, 4.6$c{m^3}$ of methyl alcohol is dissolved in 25.2g of water. Calculate
(1) % by mass of methyl alcohol
(2) Mole fraction of methyl alcohol and water. (Given density of methyl alcohol = 0.7952$g.c{m^{ - 3}}$ and C = 12, H = 1, O = 16)

Answer 1

Hint: To find % by mass of methyl alcohol we can find methyl alcohol mass by using its density. And we can easily find mole fraction of water and methyl alcohol by using mole fraction formula.

Complete answer:
(1)We should know that mass % is the mass of the element or solute divided by the mass of the compound or solute. The result is multiplied by 100 to give the %.
                  $\% mass = \dfrac{{wt.solute}}{{wt.mixture}} \times 100$ (1)
This gives us the mass percentage of methyl alcohol after it has been dissolved in water.
-The values given in the question are:
 Weight of solvent (${H_2}O$) = 25.2g
 Volume of solvent ($C{H_3}OH$) = 4.6$c{m^3}$
 Density of $C{H_3}OH$ = 0.7952$g.c{m^{ - 3}}$
 Molecular weight of $C{H_3}OH$ = 32$g.mo{l^{ - 1}}$
Now using the values of volume and density we can calculate weight of $C{H_3}OH$ present in it: $density = \dfrac{{mass}}{{volume}}$
                                $0.7952 = \dfrac{{weight}}{{4.6}}$
                                  Weight = 0.7952 × 4.6
                                                = 3.65792 g
So, the weight of the solute ($C{H_3}OH$) will be 3.65792 g.
Hence, the weight of the solution will be = weight of solute + weight of solvent
                                                                         = 3.65792 + 25.2
                                                                         = 28.85792 g
   -Now we will use the equation (1) to calculate the mass %:
                           $\% mass = \dfrac{{wt.solute}}{{wt.mixture}} \times 100$
                                              = $\dfrac{{3.65792}}{{28.85792}} \times 100$
                                              = 12.67%
So, the mass percentage of $C{H_3}OH$ will be 12.67%.
(2)Now, in the next part of the question, we have to find mole fraction of methyl alcohol and water.
Mole fraction: It is simply the number of moles of a particular substance compared to the total number of moles. We know that the number of moles of one substance in a mixture can never be more than the total number of moles, the mole fraction for a substance always ranges between 0 and 1. The mole fraction is typically given the symbol X. The mole fraction for a substance can be mathematically written as:
                                       ${X_i} = \dfrac{{{n_i}}}{{{n_{Total}}}}$ (2)
-We will first calculate the moles of ${H_2}O$ and $C{H_3}OH$:
 ${n_{{H_2}O}} = \dfrac{{{W_{{H_2}O}}}}{{{M_{{H_2}O}}}}$ = $\dfrac{{25.2}}{{18}}$ = 1.4
 ${n_{C{H_3}OH}} = \dfrac{{{W_{C{H_3}OH}}}}{{{M_{C{H_3}OH}}}}$ = $\dfrac{{3.65792}}{{32}}$ = 0.1143
    ${n_{Total}} = {n_{{H_2}O}} + {n_{C{H_3}OH}}$ = 1.4 + 0.1143 = 1.5143
-Now we will calculate the mole fraction of ${H_2}O$ using equation (2):
                         ${X_{{H_2}O}} = \dfrac{{{n_{{H_2}O}}}}{{{n_{Total}}}}$
                                                     = $\dfrac{{1.4}}{{1.5143}}$ = 0.9245
Since, ${X_{{H_2}O}} + {X_{C{H_3}OH}} = 1$; we can also write it as:
                    ${X_{C{H_3}OH}} = 1 - {X_{{H_2}O}}$
                                                     = 1 – 0.9245
                                                     = 0.0755
Hence the mole fraction of ${H_2}O$ is 0.9245 and that of $C{H_3}OH$ is 0.0755
Final answers are-
(1)The % by mass of methyl alcohol is 12.67%.
(2)Mole fraction of methyl alcohol = 0.0755 moles
     Mole fraction of water = 0.9245 moles

Note:
We should know that concentration can be expressed as percent by volume, percent by mass, and molarity. We should also know about Dilute Concentrations Units. If we have a solution, which is too dilute, their percentage concentration is too low. So, instead of using really low percentage concentrations such as 0.00001% or 0.000000001%, we choose another way to express the concentrations. We use concentration like, Parts per Million: A concentration of a solution that contains 1 g solute and 1000000 mL solution (same as 1 mg solute and 1 L solution) would create a very small percentage concentration.