Question

10 ml of 9.8% aqueous h2so4 in mixed with 10 ml of calculate the molarity of the new solution.​

Answer 1

10 percent by volume of H2SO4 means that 10 ml acid in 100 ml solution

volume of H2SO4= 10ml

Volume of water= 90 ml

Density of solution is 1.2 g/cc

mass of aq solution of H2SO4=volume*density= 1.2*100=120gm

Mass of water is= 90*1=90g

Mass of solute i.e. H2SO4= 120-90 = 30g

molality means number of moles of solute present in 1 kg solvent .

Hence molality= number of mole of solute/wt.of solvent in kg

number of mole of solute=30/98=.306

Mass of water is= 90*1=90g=.09kg

molality=.306/.09=3.40

Answer 2

Answer:

9.8% (w/w) of H2 SO4 solution means 9.8 g of solute in 100 g

of solution.

Density of H2 SO4=1.2 g/ml

The volume of the solution =

$\frac{mass}{density} = \frac{100}{1.2} = 83.33ml$

Molarity =

$\frac{9.8 \times 1000}{98 \times 83.33} = 1.2M$