States of Matter Solid, Liquid and Gas MCQ Questions & Answers in Physical Chemistry | Chemistry

When temperature is changed, water is converted to ice or steam depending upon the temperature. There is no change in the chemical properties of water and only physical state changes.

Given, $$Li$$  has a $$bcc$$  structure.
Density $$\left( \rho \right) = 530\,kg{\text{ - }}{m^{ - 3}}$$
Atomic mass $$\left( M \right) = 6.94\,g\,mo{l^{ - 1}}$$
Avogadro's number $$\left( {{N_A}} \right) = 6.02 \times {10^{23}}mo{l^{ - 1}}$$
We know that, number of atoms per unit cell in $$bcc$$  $$\left( Z \right) = 2.$$
∴ We have the formula for density,
$$\rho = \frac{{ZM}}{{{N_A}{a^3}}}$$
where $$a$$ = edge-length of a unit cell.
$$\eqalign{ & {\text{or}}\,\,a = \root 3 \of {\frac{{ZM}}{{\rho {N_A}}}} \cr & = \root 3 \of {\frac{{2 \times 6.94\,g\,mo{l^{ - 1}}}}{{0.53\,g\,c{m^{ - 3}} \times 6.02 \times {{10}^{23}}mo{l^{ - 1}}}}} \cr & = \root 3 \of {4.35 \times {{10}^{ - 23}}c{m^{ - 3}}} \cr & = 3.52 \times {10^{ - 8}}cm \cr & a = 352\,pm \cr} $$

$$\eqalign{ & {\text{According to Boyle}}'{\text{s}}\,\,{\text{law}} \cr & \frac{{{V_1}}}{{{V_2}}} = \frac{{{P_2}}}{{{P_1}}};\,\frac{{750}}{{{V_2}}} = \frac{{360}}{{840}} \cr & {V_2} = 1750\,mL = 1.750\,L \cr} $$

According to Boyle's law, $$PV=$$  constant
$$\eqalign{ & \therefore \,\,{\text{log}}\,P + {\text{log}}\,V = {\text{constant}} \cr & \,{\text{log}}\,P = - {\text{log}}\,V + {\text{constant}} \cr} $$
Hence, the plot of $${\text{log}}\,P\,\,{\text{vs}}\,\,{\text{log}}\,V$$    is straight line with negative slope.

At high pressure real gas particles are easily compressed.

$$\frac{n}{V} = \frac{P}{{RT}}$$

TIPS/Formulae :
$$\eqalign{ & {\text{Rate of diffusion}}\, \propto \,\sqrt {\frac{1}{{{\text{Molecular mass}}}}} \cr & \because \,\,{\text{Molecular mass of }}HCl{\text{ > molecular mass of}}\,N{H_3} \cr} $$
∴ $$HCl$$  diffuses at slower rate and white ammonium chloride is first formed near $$HCl$$  bottle.

Hydrogen bond is a type of strong electrostatic dipolc-dipole interaction and dependent on the inverse cube of distance between the molecular ion-dipole interaction $$ \propto \frac{1}{{{r^3}}}.$$

$$\eqalign{ & {{\text{P}}_{{N_2}}} = \frac{{0.4}}{{0.2 + 0.4 + 0.1 + 0.3}} \times 1 \cr & \,\,\,\,\,\,\,\,\,\, = 0.4\,\,atm \cr} $$
( $$\because $$  Partial pressure = mole fraction × total pressure )

Doping of $$NaCl$$   with $${10^{ - 4}}mol\,\% $$    of $$SrC{l_2}$$  means, $$100\,moles$$   of $$NaCl$$  are doped with $${10^{ - 4}}mol$$   of $$SrC{l_2}.$$
$$\therefore \,\,1\,mol$$   of $$NaCl$$  is doped with
$$SrC{l_2} = \frac{{{{10}^{ - 4}}}}{{100}} = {10^{ - 6}}\,mole$$
As each $$S{r^{2 + }}\,ion$$   introduces one cation vacancy.
$$\therefore $$  Concentration of cation vacancies
$$\eqalign{ & = {10^{ - 6}}mol/mol\,{\text{of}}\,NaCl \cr & = {10^{ - 6}} \times 6.023 \times {10^{23}}mo{l^{ - 1}} \cr & = 6.023 \times {10^{17}}mo{l^{ - 1}} \cr} $$