Question

A solution is prepared by mixing 525 mL of ethanol with 597 mL of water. The molarity of ethanol in the resulting solution is 8.35 M. The density of ethanol at this temperature is 0.7893 g/mL. Calculate the difference in volume between the total volume of water and ethanol that were mixed to prepare the solution and the actual volume of the solution. g

Answer 1

Answer:

$\Delta V = 234.736\,mL$

Explanation:

The quantity of moles of ethanol in the solution is:

$n_{C_{2}H_{5}OH} = \left(\frac{597\,mL}{1000\,mL} \right)\cdot \left(8.35\,\frac{mol}{L} \right)$

$n_{C_{2}H_{5}OH} = 4.985\,mol$

The mass and volume of ethanol in the solution are, respectively:

$m_{C_{2}H_{5}OH} = (4.985\,mol)\cdot \left(46.07\,\frac{g}{mol} \right)$

$m_{C_{2}H_{5}OH} = 229.658\,g$

$V_{C_{2}H_{5}OH} = \frac{229.658\,g}{0.7893\,\frac{g}{mL} }$

$V_{C_{2}H_{5}OH} = 290.964\,mL$

The difference between the total volume of water and ethanol mixed to prepare the solution and the actual volume of solution is:

$\Delta V = (525\,mL+597\,mL) - (597\,mL + 290.964\,mL)$

$\Delta V = 234.736\,mL$

Answer 2

The difference in volume between the total volume of water and ethanol is ΔV =234.736 mL.

Calculation for moles of ethanol:

The quantity of moles of ethanol in the solution is:

$nC_2H_5OH=\frac{597mL}{1000mL} *8.35mol/L\\\\nC_2H_5OH=4.985 moles$

The mass and volume of ethanol in the solution are, respectively:

$mC_2H_5OH=4.985moles*46.07g/mol\\\\mC_2H_5OH=229.685g$

$VC_2H_5OH=\frac{229.685g}{0.7893g/mL} \\\\VC_2H_5OH=290.964mL$

The difference between the total volume of water and ethanol mixed to prepare the solution and the actual volume of solution is:

ΔV= (525mL+597mL)- (597mL + 290.964 mL)

ΔV= 234.736mL

Find more information about Moles here:

brainly.com/question/13314627