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10

No, there is no law that requires the density of a mixture to fall between the densities of pure components. It does so most of the time, but the exceptions are not unheard of. Here's one. Water: density $1.00\ \rm{g/cm^3}$ Hydrazine: $1.02\ \rm{g/cm^3}$ Hydrazine hydrate: $1.03\ \rm{g/cm^3}$ So it goes.


5

Solution By definition, molarity $c$ is the amount of substance $n$ per volume of solution $V$: $$c = \frac{n}{V}\label{eqn:1}\tag{1}$$ The volume $V$ can be found using total mass of solution $m_\mathrm{tot}$ and its density $\rho$: $$V = \frac{m_\mathrm{tot}}{\rho}\label{eqn:2}\tag{2}$$ The amount of substance $n$ can be found using mass of solute $m$ and ...


2

Your mistake is that you have calculated the mass of a sample with the same volume as ours consisting entirely of sand, and a sample with the same volume as ours consisting entirely of sand. volume = mass_total / density_total = 0.2 kg / 2300 kg/m^3 = 0.00008696 m^3 gives the initial volume of the sample before the salt is removed, however by multiplying ...


1

a) molar concentration There are indeed 6.14 moles of $\ce{NaCl}$ in 359 grams of $\ce{NaCl}$. I'd prefer to write the molar mass as "$\pu{58.44 g mol^-1 }\ \ce{NaCl}$" or "$\pu{58.44 g/mol}\ \ce{NaCl}$" to keep it all together. b) molal concentration You got the right value, but you did it the wrong way. For the solvent: $$\dfrac{\pu{980 ...


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