Question

An antifreeze solution is prepared from 333.2 g of ethylene glycol and 200 g of water. calculate the molarity of the solution.if the density of the solution is 1.072gml-1, then what shall be the molarity of the solution?​

Answer 1

C2H6O2= 62

density=mass/volume

or,v=m/d

= 200/1.072

=186.56cm3

= 0.186 L

so...M=( 333.2/62)/(1/0.186)

=0.9996 mol/L

Hope it helps

Answer 2

$\fbox{\texttt{\green{Answer:}}}$

Molality of the solution = $\bold{17.95\:mol\:kg^{-1}}$

Molarity of the solution = $\bold{9.11\:mol\:L^{-1}}$

$\fbox{\texttt{\pink{Given:}}}$

Mass of the solute $\bold{C_2H_4(OH)_2}$ = 222.6 g

Mass of the solvent (water) = 200 g = 0.200 kg

Density of the solution = $\bold{1.072\:g\:mL^{-1}}$

$\fbox{\texttt{\orange{To\:find:}}}$

Molality & Molarity of the solution.

$\fbox{\texttt{\red{How\:to\:Find:}}}$

Molar mass of $\bold{C_2H_4(OH)_2=62\:g\:mol^{-1}}$

Also, mass of solute = 222.6 g (given).

$\therefore$ Moles of solute = $\bold{\frac{222.6\:g}{62\:g\:mol^{-1}}=3.59}$

We know, mass of the solvent = 0.200 kg (given).

So, Total mass of the solution = 422.6 g

Then, volume of the solution = $\bold{\frac{422.6\:g}{1.072\:g\:mL^{-1}}}$ = 394.2 mL = 0.3942 L

$\hrulefill$

Applying the formula,

$\bold{Molality=\frac{Moles\:of\:the\:solute}{Mass\:of\:the\:solvent}}$

We get, Molality =

$\bold{\frac{3.59\:moles}{0.200\:kg}=17.95\:mol\:kg^{-1}}$

$\hrulefill$

Applying the formula,

$\bold{Molarity=\frac{Moles\:of\:the\:solute}{Volume\:of\:the\:solution}}$

We get, Molarity =

$\bold{\frac{3.59\:moles}{0.3942\:L}=9.11\:mol\:L^{-1}}$