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Ivahew [28]
3 years ago
6

For an alloy that consists of 78.2 g copper, 103.5 g zinc, and 2.8 g lead, what are the concentrations of (a) Cu, (b) Zn, and (c

) Pb in weight percent? The atomic weights of Cu, Zn, and Pb are 63.54, 65.39, and 207.2 g/mol, respectively.
Chemistry
1 answer:
TEA [102]3 years ago
8 0

Answer:

Explanation:

Mass percent is defined as the mass of an element divided by the sum of masses of all the elements multiplied by 100. It is generally used to define the concentration. It does not depend on concentration.

It is given as;

Mass percent = (mass of an element / Total mass of the compound ) × 100

Mass of  compound (alloy) = 78.2 g copper  + 103.5 g zinc +2.8 g lead = 184.5 g

(a) Cu

Mass of Cu = 78.2

Mass percent = (78.2 / 184.5) * 100 = 42.38%

(a) Zn

Mass of Zn = 103.5

Mass percent = (103.5 / 184.5) * 100 = 56.10%

(c) Pb

Mass of Pb = 2.8

Mass percent = (2.8 / 184.5) * 100 = 1.52%

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What is the mole fraction, X, of solute and the molality, m (or b), for an aqueous solution that is 15.0% NaOH by mass?
GalinKa [24]
Hello!

a) The mole fraction of solute of a 15% NaOH aqueous solution can be calculated in the following way:

First, we have to assume that we have 100 grams of solution. This will simplify the calculations.

Now, we know that this solution has 15 grams of NaOH and 85 grams of water. We can calculate the number of moles of each one in the following way:

molesNaOH= 15 g NaOH*\frac{1molNaOH}{39,997 g NaOH}=0,3750 moles NaOH \\ \\ moles H_2O=85 gH_2O* \frac{1 mol H_2O}{18 g H_2O}=4,7222 moles H_2O

To finish, we calculate the mole fraction by dividing the moles of NaOH between the total moles:

X_{NaOH}= \frac{moles NaOH}{total moles}= \frac{0,3750 moles NaOH}{0,3750 moles NaOH+4,7222 molesH_2O} =0,073

So, the mole fraction of NaOH is 0,073

b) The molality (moles NaOH/ kg of solvent) of a 15% NaOH aqueous solution can be calculated in the following way:

First, we have to assume that we have 100 grams of solution. This will simplify the calculations.

Now, we know that this solution has 15 grams of NaOH and 85 grams (0,085 kg) of water. We can calculate the moles of NaOH in the following way:

molesNaOH= 15 g NaOH*\frac{1molNaOH}{39,997 g NaOH}=0,3750 moles NaOH

Now, we apply the definition of molality to calculate the molality of the solution:

mNaOH= \frac{moles NaOH}{kg_{solvent}}=  \frac{0,3750 moles NaOH}{0,085 kg H_2O}=4,41 m

So, the molality of this solution is 4,41 m

Have a nice day!
4 0
3 years ago
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