Calculate the mole fraction of each solute and solvent: (a)...

Calculate the mole fraction of each solute and solvent: (a) $0.710 \mathrm{kg}$ of sodium carbonate (washing soda), $\mathrm{Na}_{2} \mathrm{CO}_{3},$ in $10.0 \mathrm{kg}$ of water $-$ a saturated solution at $0^{\circ} \mathrm{C}$ (b) $125 \mathrm{g}$ of $\mathrm{NH}_{4} \mathrm{NO}_{3}$ in $275 \mathrm{g}$ of water $-$ a mixture used to make an instant ice pack (c) 25 g of $\mathrm{Cl}_{2}$ in $125 \mathrm{g}$ of dichloromethane, $\mathrm{CH}_{2} \mathrm{Cl}_{2}$ (d) 0.372 g of tetrahydropyridine, C $_{5} \mathrm{H}_{9} \mathrm{N},$ in $125 \mathrm{g}$ of chloroform, $\mathrm{CHCl}_{3}$

Calculate the mole fraction of each solute and solvent:
(a) $0.710 \mathrm{kg}$ of sodium carbonate (washing soda), $\mathrm{Na}_{2} \mathrm{CO}_{3},$ in $10.0 \mathrm{kg}$ of water $-$ a saturated solution at $0^{\circ} \mathrm{C}$
(b) $125 \mathrm{g}$ of $\mathrm{NH}_{4} \mathrm{NO}_{3}$ in $275 \mathrm{g}$ of water $-$ a mixture used to make an instant ice pack
(c) 25 g of $\mathrm{Cl}_{2}$ in $125 \mathrm{g}$ of dichloromethane, $\mathrm{CH}_{2} \mathrm{Cl}_{2}$
(d) 0.372 g of tetrahydropyridine, C $_{5} \mathrm{H}_{9} \mathrm{N},$ in $125 \mathrm{g}$ of chloroform, $\mathrm{CHCl}_{3}$

Key Concepts

Solution Composition

Solution composition refers to the makeup of a solution, which includes both the solute(s) and the solvent. Understanding the individual contributions of each component—in terms of moles—is key for calculating properties like mole fraction. This concept is foundational in solution chemistry, where the interactions and proportions of different substances affect the overall behavior and properties of the solution.

Mole Fraction

Mole fraction is a way to express the concentration of a component in a mixture. It is defined as the ratio of the number of moles of the component to the total number of moles of all components present in the mixture. This dimensionless quantity is particularly useful in thermodynamic calculations and provides a way to compare the relative amounts of substances regardless of their molar masses.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It serves as a conversion factor between the mass of a substance and the number of moles, enabling calculations that relate the mass of a chemical to the number of its constituent particles. Accurate molar mass values are essential for computations involving mole fractions, stoichiometry, and reaction yields.

Mass-to-Mole Conversion

Mass-to-mole conversion is the process of converting a given mass of a substance to the number of moles using its molar mass. This conversion is crucial in chemical calculations because reactions occur based on mole ratios rather than mass ratios. The conversion is performed by dividing the mass of the substance by its molar mass.

Transcript

0:00 Now we get ready.

00:01 In this problem we are asked to calculate the mole fraction and we're doing for the solute and the solvent.

00:10 And we have four scenarios that we're going to do.

00:13 Our first 0 .710 kilograms of sodium carbonate in 10 .0 kilograms of h2o.

00:29 And we're told that this is a saturated solution at zero degrees c um really all that pertinent but our molar mass of these and let me get to my lentech molar mass solver if i got one open here i do not what are the odds of that i'm a big fan of lentech molar mass so na2co3 is 105 .99 na2co3 and of course water is 18 .02 grams per mole.

01:21 So my mole fraction which is represented by x is going to be moles of solute divided by total moles and then we'll change that should be an l to solvent.

01:44 So first i'm going to set this up as one big problem here you can set it up in smaller pieces if you choose.

01:56 Since these are both in kilograms.

01:58 I wouldn't have to convert this, but i'll go ahead and convert it.

02:02 It's not a big deal.

02:03 I usually just do this conversion in my head.

02:06 And then i'm going to multiply that by one, the reciprocal of the molar mass.

02:15 And for my denominator, pretty simple here, i'm going to have the sum of 0 .710 kilograms times, in fact, this one i'm going to cheat a little bit, and i'm going to write 710 grams times 105 .99 grams per mole.

02:43 Plus, here i have 10 ,000, significant to here, grams times 18 .02 grams per mole.

03:00 Okay, let's do this.

03:02 0 .71 times 1000 divided by 105 .99.

03:12 So here i'm getting my mole fraction of my na2co3 will equal 6 .6987 divided by 6 .6987 plus 10 ,000, whoops, too many zeros, divided by 18 .02.

03:41 Usually we have a pretty low fraction, or often we have a pretty low fraction, fraction, mole fraction of our solute, 554 .939.

03:53 And this won't have any units.

03:56 So i'm going to take 6 .6987 divided by parentheses 6 .6987 plus 554 .939.

04:08 Close.

04:11 And i get 0 .0119.

04:16 Nine.

04:16 Let me check my sig figs on this one.

04:22 Three and three.

04:24 So i'm going to be good with that.

04:26 And my mole fraction of my h2o will simply be one minus 0 .0119.

04:35 And that will equal equal.

04:36 1 minus second answer, enter.