Question

molarity of H2SO4 is 0.8 and its density is 1.06gcm-3 what will be the concentration of the of the solution in rerins of molality and mole fraction ​

Answer 1

$\huge{\mathfrak{Question:-}}$

Molality of H2SO4 is 0.8 and its density is 1.06 gcm^-3 what will be the concentration of the of the solution in terms of molality and mole fraction.

$\huge{\mathfrak{Solution:-}}$

Since molality of the $\sf{H_{2}SO_{4}}$ solution is given as 0.8 M. It means that 0.8 moles of$\sf{H_{2}SO_{4}}$ are present in 1000 ml or 1000 cm³ of the solution.

The mass of the $\sf{H_{2}SO_{4}}$ dissolved can be found out as  $\sf{0.8\times 98=78.4\;g.}$

Again density of the solution (1.06 g/cm³) is given.

Hence we find mass of the solution as follows:

$\sf{Mass\;of\;solution=1.06\times 1000 = 1060\;g}$

$\sf{Thus\;mass\;of\;water\;only=(1060-78.4)=981.6\;g}$

$\sf{Thus\;moles\;of\;water=\frac{918.6}{18}=54.4\;moles}$

To find the molality, we have,

$\sf{Molality=\big(\frac{0.8}{981.6}\big)\times 1000=0.81\;m}$

$\sf{To\;find\;the\;mole\;fraction\;of\;H_{2}SO_{4}\;(x_{1})\;and\;water\;(x_{2}),\;we\;have}$

$\sf{x_{1}=\frac{0.8}{(0.8+54.5)}=0.014}$

$\sf{x_{2}=(1-0.014)=0.98}$

Answer 2

$\huge{\mathbf{\red{\underline{\underline{Solution:-}}}}}$

Molarity of a solution is 0.8 M, it means the solution have 0.8 moles of H2SO4 in 1000 ml of solution.

Mass of 1000 ml of solution = density × volume.

= 1.06 g cm^3 × 1000 cm^3

= 1060 g.

Mass of H2SO4 in 0.8 mol of H2SO4 = 0.8 mol × 98 g mol

= 78.4 g

Mass of water (Solvent) = 1060 g - 78.4 g

= 981.6 g

Moles of water (Solvent) = 981.6 g / 18 g mol

= 54.53 mol.

Since, molality is the moles of solute in 1 kg of the solvent, the molality of the solution is = ( 0.8 mol / 981.6 g ) × 1000 g

= 0.815 m

Mole fraction of H2SO4 = (0.8) / (0.8 + 54.53)

= 0.0145

Mole fraction of water = (54.53) / (0.8 + 54.53)

= 0.985.