# Tag Info

Water is very unique liquid, because it has a higher density in the liquid state than the solid state. The maximum density of water is found at $\pu{4 ^\circ C}$, which is reported as $\pu{999.9720 kg\:m^{-3}}$ (Temperature Effects on Density). Therefore, molarity of water at $\pu{4 ^\circ C}$ can be calculated as $\frac{\pu{999.9720 g\:L^{-1}}}{\pu{18.015 g\... 1 I think you have confused amount of substance with molarity of the solution. For example:$\ce{NaOH}$solution with$\pu{1M}$in molarity means$\pu{1L}$of solution contains$\pu{1mol}$of$\ce{NaOH}$. You can calculate amount of$\ce{NaOH}$in any volume of the solution by multiplying its molarity and required volume in$\pu{L}: $$\text{Amount of }\ce{... 0 There is problem with ppx values they are ambiguous. It may be w/w, w/v, v/v, n/n. Salt water has density significantly different to \pu{1 g/ml}, so \pu{1 ppt(parts per thausand) } may mean \pu{1000 mg/L} or \pu{1000 mg/kg}, with the recalculation factor of the solution density. The former (\pu{ppt w/v as 1000 mg/L}) is more probable, but check ... 1 First, when solving problems like this one, you have to make sure that both salts possess good solubility and that no double salt is precipitating:$$ \begin{align} \ce{NaNO3 &<=>> Na+ + NO3-}\label{rxn:R1}\tag{R1}\\ \ce{Ca(NO3)2 &<=>> Ca^2+ + 2NO3-}\label{rxn:R2}\tag{R2} \end{align} $$Both nitrates are indeed well soluble in ... 2$$\pu{1 mg L-1} = \pu{1 ppm} = \frac{1}{1000}\cdot\pu{ppt}$$or$$\pu{1 ppt} = \pu{1000 ppm}$$For example:$$\pu{10 ppm} = \frac{10}{1000}~\pu{ppt} = \pu{0.01 ppt}\tag{1}\pu{5 ppt} = 5\cdot\pu{1000 ppm} = \pu{5000 ppm}\tag{2}\$