A solution of H 2 SO 4 ( aq ) H2SO4(aq) with a molal concentration of 8.01 m 8.01 m has a density of 1.354 g / mL . 1.354 g/mL. What is the molar concentration of this solution?

[H₂SO₄] = 6.07 M

Explanation:

Analyse the data given

8.01 m → 8.01 moles of solute in 1kg of solvent.

1.354 g/mL → Solution density

We convert the moles of solute to mass → 8.01 mol . 98g /1mol = 785.4 g

Mass of solvent = 1kg = 1000 g

Mass of solution = 1000g + 785.4 g = 1785.4 g

We apply density to determine the volume of solution

Density = Mass / volume → Volume = mass / density

1785.4 g / 1.354 g/mL = 1318.6 mL

We need this volume in L, in order to reach molarity:

1318.6 mL . 1L / 1000mL = 1.3186 L ≅ 1.32L

Molarity (mol/L) → 8.01 mol / 1.32L = 6.07M

6.08M

Explanation:

8.010 m means 8.010 mol / 1 kg of solvent

Molar mass of H2SO4=98g/mol

8.01mol × 98g/mol = 784.98 g of solute

784.98g + 1000 g = 1784.98g total for solute and solvent in the 8.010 m solution.

But density of acid=1.354g/ml

1784.98 g ÷ 1.354 g/mL = 1318.30 mL

1318.30/1000 = 1.3183L

8.01 moles / 1.3183 L = 6.08 M

6.08M

We will call D1 and D2 the volume of dextrose at 2.5% and 30% respectively

D1 + D2 = 500 (1)

D1*2,5/100 + D2*30/100 = 500*10/100 (2)

(2,5D1+30D2)/100 = 50

2,5D1 +30D2 = 50*100 = 5.000 (2)

D1 = 500 - D2 Now we clear D1 in (1)

2,5(500-D2) + 30D2 = 5.000 and we substitute in (2)

1.250 - 2,5D2 + 30D2 = 5.000

27,5D2 = 5.000 - 1.250 = 3.750

D2 = 3.750/27,5 = 136ml +4/11ml

D1 = 500ml - 136ml +4/11ml

D1 = 363ml +7/11ml

ANSWER 363ml+7/11 ml of dextrose 2,5% and 136ml+4/11 ml dextrose 30%