Concentrations: Dilution and mixture

Tutorial 3

        

Progressive exercise What volume of concentrated sulfuric acid of density $ 1.84 $ containing $ 98 \% $ of $ H_2SO_4 $ must be introduced into the water to make $ 200 \; cm^3 $ of a $ 0.25 \; M $ solution? Calculation: Final solution: First we calculate .......... the number of sulfuric acid moles:
$n_{so}=0,25\cdot 0,200=0,050\; mol $ To continue we need ..........the mass of sulfuric acid:
$m_{so}=0,050\cdot 98= 4,9 \;g$ Concentrated solution:: We now know ..........the mass of sulfuric acid: $4,9\; g$ and its percentage: $98 \;\%$ It is therefore possible to calculate ..........the mass of the concentrated solution:
$m_S=\frac{4,9\cdot 100}{98}$ = $5,0\;g$ The density of the concentrated solution is: ..........$\rho_S=1,84 \frac{g}{mL}$ which leads to its volume: ..........$V_S=\frac{m_S}{\rho_S}=\frac{5,0}{1,84}= 2,72 \;mL$