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masya89 [10]
3 years ago
11

Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040(assume the density of wa

ter to be one).
Chemistry
1 answer:
andreev551 [17]3 years ago
8 0
  As the given  <span>moles OF ethanol = 0.040 
now will find water moles that will b
moles of water = 1 - 0.040=0.96 (substract 1 from etahnol mole fraction)
now the
mass of water = 0.96 mol x 18.02 g/mol=17.3 g  where 18 is atomic mass of water
mass of ethanol = 0.040 x 46 g/mol =1.84 g 
 now
mass of solution = 17.3 + 1.84=19.1 g  that is mass of water +mass of etanol
Now
assume that the density of the solution = 1g/mL 
so the
volume  ofsolution = 19.1 mL = 0.0191 L  
so finally the molarity will be
M = 0.040/ 0.0191=2.1
hope it helps</span>
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specific heat for the metal is given as 0.444 J per g per degree C.

let's calculate the heat lost by iron metal using the equation:

q=mc\Delta T

where, q is the heat energy, m is mass, c is specific heat and delta T is change in temperature. let's plug in the values and calculate q for iron metal:

q=93.3g(45.9^0C)(\frac{0.444J}{g.^0C})

q = 1901.42 J

Using same equation we will calculate the heat gained by water.

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specific heat for water is 4.184 J pr g per degree C. Let's plug in the values:

q=75.0g(\frac{4.184J}{g.^0C})(2.73^0C)

q = 856.674 J

Total heat lost by iron metal is the sum of heat gained by water and calorimeter.

So, heat gained by calorimeter = heat lost by iron metal - geat gained by water

heat gained by calorimeter = 1901.42 J - 856.674 J = 1044.746 J

Change in temperature for calrimeter is same as for water that is 2.73 degree C

For calorimeter, q=C.\Delta T

C=\frac{q}{\Delta T}

C=\frac{1044.746J}{2.73^0C}

C=\frac{382.69J}{0C}

So, the heat capacity of calorimeter is \frac{382.69J}{0C} .


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