Question

The concentrated sulfuric acid we use in the laboratory is 98% H_2SO_4 by mass. Calculate the molality and molarity of the acid solution. The density of the solution is 1.83 \frac{g}{mL} answers are 5\times10^{-2} m, 18.3M

Other
ANSWERED
asked 2020-11-02
The concentrated sulfuric acid we use in the laboratory is 98% \(\displaystyle{H}_{{2}}{S}{O}_{{4}}\) by mass. Calculate the molality and molarity of the acid solution. The density of the solution is 1.83 \(\displaystyle{\frac{{{g}}}{{{m}{L}}}}\) answers are \(\displaystyle{5}\times{10}^{{-{2}}}\) m, 18.3M

Answers (1)

2020-11-03
Mass percent of \(\displaystyle{H}_{{2}}{S}{O}_{{4}}\) solution=98%
98%(m/m) \(\displaystyle{H}_{{2}}{S}{O}_{{4}}\) solution means 98 g of \(\displaystyle{H}_{{2}}{S}{O}_{{4}}\) solute is present per 100 g of solution.
Mass of \(\displaystyle{H}_{{2}}{S}{O}_{{4}}\) solute =98 g
Mass of water =(100-98)g
\(\displaystyle={2}{g}\times{\frac{{{1}\ {k}{g}}}{{{1000}\ {g}}}}\)
=0.002 kg
Mass of \(\displaystyle{H}_{{2}}{S}{O}_{{4}}\) solute =98 g
Molar mass of \(\displaystyle{H}_{{2}}{S}{O}_{{4}}\) solute =98.09 g/mol
No of moles of \(\displaystyle{H}_{{2}}{S}{O}_{{4}}\) solute \(\displaystyle={98}{g}\times{\frac{{{1}{m}{o}{l}}}{{{98.09}{g}}}}\)
=0.999 mol
\(\displaystyle\text{Molality}={\frac{{\text{No of moles of }\ {H}_{{2}}{S}{O}_{{4}}\ \text{ solute}}}{{\text{mass of }\ {H}_{{2}}{O}\ \text{ solvent}}}}\)
\(\displaystyle={\frac{{{0.999}\ {m}{o}{l}}}{{{0.002}\ {k}{g}}}}\)
\(\displaystyle={5.0}\times{10}^{{2}}\) mol/kg
\(\displaystyle={5.0}\times{10}^{{2}}\) m
Therefore, the molality of the 98% \(\displaystyle{H}_{{2}}{S}{O}_{{4}}\) solution is \(\displaystyle{5.0}\times{10}^{{2}}\) m
Mass of \(\displaystyle{H}_{{2}}{S}{O}_{{4}}\) solution=100g
Density of \(\displaystyle{H}_{{2}}{S}{O}_{{4}}\) solution=1.83\(\displaystyle{\frac{{{g}}}{{{m}{L}}}}\)
Volume of solution \(\displaystyle={100}\ {g}\ {H}_{{2}}{S}{O}_{{4}}\times{\frac{{{1}\ {m}{L}}}{{{1.83}\ {g}\ {H}_{{2}}{S}{O}_{{4}}}}}\times{\frac{{{1}\ {L}}}{{{1000}\ {m}{L}}}}\)
=0.999 mol
\(\displaystyle\text{Molality}={\frac{{\text{No of moles of }\ {H}_{{2}}{S}{O}_{{4}}\ \text{ solute}}}{{\text{Volume of solution}}}}\)
\(\displaystyle={\frac{{{0.999}\ {m}{o}{l}}}{{{0.0546}\ {L}}}}\)
=18.3 mol/L
=18.3 M
Therefore, the molarity of \(\displaystyle{H}_{{2}}{S}{O}_{{4}}\) solution is 18.3 M
0
 
Best answer

expert advice

Have a similar question?
We can deal with it in 3 hours

Relevant Questions

asked 2021-02-14
Mg reacts with H+ (aq) according to
\(\displaystyle{M}{g{{\left({s}\right)}}}+{2}{H}+{\left({a}{q}\right)}\to{M}{g}_{{2}}+{\left({a}{q}\right)}+{H}_{{2}}{\left({g}\right)}\)
Suppose that 0.524 g of Mg is reacted with 60.o ml of 1.0 M H+(aq). Assume that the density of the H+ (aq) solution os 1.00 g/ml,and that its specific heat capacity equals that of water. Theinitial and final temperatures are 22.0 degree celsius and 65.8degree celsius.
a) Is the reaction endothermic or exothermic?
b) Calculate \(\displaystyle\triangle{H}\) of of th reaction. Use correct sigs andgive units.
c) Calculate the \(\displaystyle\triangle{H}\) of the reaction per mole of magnesium.
asked 2021-04-19
In the figure, a cube of edge length L = 0.599 m and mass 970 kg is suspended by a rope in an open tank of liquid of density 1.05E+3 kg/m3. Find (a) the magnitude of the total downward force on the top of the cube from the liquid and the atmosphere, assuming atmospheric pressure is 1.00 atm, (b) the magnitude of the total upward force on the bottom of the cube, and (c) the tension in the rope. (d) Calculate the magnitude of the buoyant force on the cube using Archimede's principle.
asked 2021-03-26
Your can dissolve an aluminum soft drink can in an(aq) base such as potassium hydroxide.
\(\displaystyle{2}{A}{l}{\left({s}\right)}+{2}{K}{O}{H}{\left({a}{q}\right)}+{6}{H}_{{2}}{O}\to{2}{K}{A}{l}{\left({O}{H}\right)}_{{4}}{\left({a}{q}\right)}+{3}{H}_{{2}}{\left({g}\right)}\)
If you place 2.05 g of alumimum in a beaker with 185 mL of 1.35 M KOH, will any aluminum remain(show all calculations). What mass of \(\displaystyle{K}{A}{l}{\left({O}{H}\right)}_{{4}}\) is produced?
asked 2021-05-12
Does any solid \(\displaystyle{A}{g}_{{2}}{C}{r}{O}_{{4}}\) form when \(\displaystyle{2.7}\times{10}^{{-{5}}}\) g of \(\displaystyle{A}{g}{N}{O}_{{3}}\) is dissolved in 15.0 ml of \(\displaystyle{4.0}\times{10}^{{-{4}}}\ {K}_{{2}}{C}{r}{O}_{{4}}\)?
...