Question

Determine the molarity of antifreeze solution containing 250g water mixed with 222g ethylene glycol. The density of solution is 1.07g/ml

Answer 1

Answer:

Explanation:

No. of moles of ethylene glycol = nB Molar mass of C2H6O2 = (24 + 6 + 32)g mol–1 NB = 222g/62mol-1 = 3.58mol mass of solution = (250 + 222)g = 472gRead more on Sarthaks.com - https://www.sarthaks.com/71616/determine-molarity-antifreeze-solution-containing-water-mixed-ethylene-glycol-c2h6o2

Answer 2

The molarity of antifreeze solution containing 250g water mixed with 222g ethylene glycol is 8.12 M

Explanation:

Molarity is defined as the number of moles of solute dissolved per Liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n= moles of solute  

$V_s$ = volume of solution in ml = 1000 ml

Given : mass of solute (ethylene glycol) = 222 g

mass of solvent (water) = 250 g

mass of solution = mass of solute + mass of solvent = (222 + 250) g= 472 g

Density of solution = 1.07 g/ml

Volume of solution =$\frac{\text {mass of solution}}{\text {density of solution}}=\frac{472g}{1.07g/ml}=441ml$

moles of solute (ethylene glycol)=$\frac{\text {given mass}}{\times {\text {molar mass}}=\frac{222g}{62g/mol}=3.58moles$

Now put all the given values in the formula of molarity, we get

$Molarity=\frac{3.58moles\times 1000}{441ml}=8.12mole/L$

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