IB Chemistry S1.1: Introduction to the particulate nature of matter worked solutions

Only one element occurs. Bonding four atoms into each particle does not introduce another kind of atom. Therefore C is correct.

In vessel I, Z is dispersed in water and takes (aq). In vessel II, the pure melt is liquid and takes (l). Therefore A is correct.

Each bead is a solid object. Moving whole beads does not free their internal particles from their local positions; they still vibrate. Therefore A is correct.

The non-reactive solvent separates two chemically different original substances. The powder therefore contained more than one substance, so B is correct.

Filter paper retains the insoluble oxide as residue. Dissolved potassium nitrate passes through in the filtrate, which is the desired liquid. Therefore A is correct.

L travels less because it spends a larger fraction of the run associated with the stationary phase. Molecular mass alone does not determine migration. Therefore D is correct.

Constant temperature fixes the model mean kinetic energy. Absorbed energy raises potential energy as the solid structure changes during melting. Therefore D is correct.

The tests explicitly preserve the molecules. Only their arrangement and spacing change when the solid sublimes. Therefore C is correct.

The stated change is directly from solid to gas, so it is sublimation. Separating particles against attractions requires energy uptake. Therefore D is correct.

Convert before taking the ratio: T_i=126.85+273.15=400.00\,\mathrm{K} and T_f=26.85+273.15=300.00\,\mathrm{K}. The mean-energy ratio is 300.00/400.00=0.750, so C is correct.

(a) It is a compound [1], containing different elements chemically joined at the fixed A:B ratio 1:2 [1]. (b) The direct gas-to-solid transition is deposition [1]. The sample releases energy to the surroundings [1].

(a) Gas particles move randomly and collide [1]. Initially more coloured particles leave the concentrated region than enter it, producing net diffusion until the composition becomes uniform [1]. (b) Gas particles are widely spaced, allowing gaps to decrease under compression [1]. The solid wall has closely packed particles held near local positions, so it retains shape [1].

(a) J boils at the specified conditions [1] because its equilibrium vapour pressure at 70 degrees Celsius is equal to the external pressure, 100 kPa [1]. (b) Its boiling temperature decreases [1]. J already has vapour pressure 72 kPa at 55 degrees Celsius, so this lower temperature meets the new boiling condition [1].

(a) Solid and liquid coexist during the specified melting interval [1]. Molecules retain the same chemical identity in both phases [1]. (b) Constant temperature means unchanged mean particle kinetic energy [1]. Input energy instead increases potential energy as the solid arrangement changes against intermolecular attractions; it does not break covalent bonds inside molecules [1].

(a) The ratio of model mean energies equals the kelvin ratio: 240.0/T_i=0.800 [1], so T_i=300\,\mathrm{K} [1]. (b) Subtract the scale offset: 300-273.15=26.85\,{}^{\circ}\mathrm{C} [1], giving 26.9\,{}^{\circ}\mathrm{C} to one decimal place [1].

(a) The layered cell is heterogeneous because its interface separates regions of different composition [1]. The uniformly sampled cell is homogeneous because top and bottom samples have the same composition [1]. (b) A homogeneous liquid can contain several substances [1]. Sampling alone does not establish a single bonded substance with definite composition [1].

(a) The dye is the solute because it is the material dissolved [1]. Propanone is the solvent, the dissolving medium [1]. The complete uniform mixture is the solution [1]. (b) The solvent is propanone rather than water, so aqueous notation does not apply [1]. (c) Total solution mass is 20.60 g [1]. Percentage dye = (0.60/20.60) times 100 = 2.91% [1]. A 20.0 g denominator compares dye mass with solvent mass rather than with solution mass [1].

(a) Heat the rinse so that the volatile solvent forms vapour [1]. Pass this vapour through a cooled condenser and collect the resulting liquid in a receiver [1]. The non-volatile resin remains in the original flask and is recovered after suitable drying [1]. (b) Place the bulb in the vapour stream at the side-arm entrance [1]. (c) A reflux condenser cools the vapour and returns liquid to the original flask [1], so no separate receiver collects the solvent [1].

(a) Choose ethanol [1]. It dissolves Q while P remains solid [1]. Filter and retain P as the residue [1]. (b) Add water to dissolve P; filter out R and keep the filtrate [1]. Evaporate the water and dry the remaining P [1]. (c) P and S both dissolve and enter the filtrate [1]. Evaporation then leaves both, so the residue is not pure P [1].

(a) Hot capacity =32.0\times40.0/100=12.8\,\mathrm{g}, sufficient for all 8.00 g U [1]. Cold dissolved U =4.00\times40.0/100=1.60\,\mathrm{g} [1]. Ideal crystal mass =8.00-1.60=6.40\,\mathrm{g} [1]. (b) Separate the crystals from the mother liquor by filtration [1], then rinse with a small cold portion and dry; do not evaporate the impurity-containing mother liquor onto the product [1]. (c) Heat the separated liquid to vaporise solvent [1]; cool the vapour in a condenser and collect the distillate [1]. (d) Minimum hot volume is (8.00/32.0) times 100 = 25.0 cubic centimetres [1]. At 20.0 cubic centimetres the hot capacity is 6.40 g [1], less than 8.00 g, so complete dissolution fails and the purification assumption is lost [1]. (e) The product melts at a lower temperature and over a wider interval than the reference, supporting residual impurity [1]. Recovery mass alone cannot establish composition, and melting observations do not conclusively prove purity or identity [1].

(a) The temperatures are 450.00 and 300.00\,\mathrm{K} [1]. The mean kinetic-energy ratio is 300/450=0.667 [1]. With the supplied relation, the rms-speed ratio is \sqrt{300/450}=0.816 [1]. (b) Energy is released from the freezing sample [1] as its potential energy decreases while its structure changes [1]. Its temperature and mean kinetic energy remain constant; molecular motion, including vibrations, continues [1]. (c) Equal absolute temperatures give equal mean translational energies [1]. Using the kinetic-energy relation, the rms-speed ratio v(H)/v(L) is the square root of m(L)/m(H) [1], giving 0.500 [1]. The greater mass is offset by a lower mean squared speed, so the proposed doubled mean energy is incorrect [1].

(a) X has the higher final mean kinetic energy because its final temperature is higher [1]. (b) X loses 0.003\,\mathrm{g} more mass but cools by 0.0\,{}^{\circ}\mathrm{C}, while Y cools by 3.5\,{}^{\circ}\mathrm{C} [1]. These results do not support the claimed necessary relation [1]. (c) Energetic surface particles escape, tending to reduce the remaining mean energy [1]. Energy supplied by the support or surroundings can replace that loss in X, keeping its temperature constant [1].

(a) Melting requires 30.0(334)=10020\,\mathrm{J} [1], below the supplied 12000\,\mathrm{J}, so all ice melts [1]. Energy left =12000-10020=1980\,\mathrm{J} [1]. Then \Delta T=1980/(30.0\times4.18)=15.8\,\mathrm{K}, giving 15.8\,{}^{\circ}\mathrm{C} [1]. (b) During melting, energy raises potential energy as the intermolecular arrangement changes while mean kinetic energy is unchanged [1]. Afterwards the liquid warms and its mean kinetic energy rises [1]. (c) Hydrogen-bonded ice has an open intermolecular network [1]. Partial collapse allows closer packing at the same mass; the water molecules and their O–H bonds remain intact [1].

(a) Convert 100\,\mathrm{kPa} to 1.00\times10^5\,\mathrm{Pa} and rearrange V=nRT/P [1]. Initially V=0.150(8.31)(300)/10^5=0.0037395\,\mathrm{m^3}=3.74\,\mathrm{dm^3} [1]. Finally V=0.0044874\,\mathrm{m^3}=4.49\,\mathrm{dm^3} [1]. (b) Mean-energy factor =360/300=1.20 [1]. In this model the mean translational energy per particle depends on absolute temperature, not molecular mass [1]. (c) Warming increases typical speeds and collision momentum transfers [1]. Expansion reduces wall-collision frequency per unit area enough to keep pressure constant [1]. Finite particle volumes or appreciable attractions invalidate ideal assumptions, especially at high density or low temperature [1].

(a) N has the higher equilibrium vapour pressure [1] because weaker attractions allow more particles to escape at a given temperature [1]. (b) Required energy =2.00(850)=1700\,\mathrm{J}=1.70\,\mathrm{kJ} [1]. Deficit =1.70-1.20=0.50\,\mathrm{kJ} [1]; internal energy of the liquid supplies the balance, so its temperature falls [1]. (c) Preferential loss of energetic surface particles lowers the remaining mean energy when input does not replace the energy loss [1]. Route vapour through a cooled condenser [1] and collect condensed liquid in a receiver, maintaining a pressure-relief path [1].