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Goshia [24]
3 years ago
6

The molarity of a solution prepared by dissolving 17.0 g of hydrochloric acid (HCl (aq)) in 133 mL of water is ___ M (2 decimal

place).
Chemistry
1 answer:
Lera25 [3.4K]3 years ago
7 0

The molarity is 3.50 M/L.

Explanation:

Molarity is found to know the amount of the solute ions present in the given volume of solution. So we determine it using the ratio of the moles of solute to the volume of the solution.

As here the weight of the solute which is hydrochloric acid is given as 17g, so we need to find the moles of it. This can be done by dividing the given amount of HCl with the molecular weight of HCl.

Molecular weight of HCl = 1.01 + 35.45 = 36.46 g/mol

So the number of moles of HCl present here will be

No.of moles = \frac{17}{36.46} =0.466

So, 0.466 moles of HCl is present in the solution.

Molarity = \frac{No.of moles of solute}{Volume of solution} = \frac{0.466}{133*10^{-3} }

Molarity = 3.50 M/L

Thus, the molarity is 3.50 M/L.

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What is the concentration of x2??? in a 0.150 m solution of the diprotic acid h2x? for h2x, ka1=4.5??10???6 and ka2=1.2??10???11
lutik1710 [3]
The first dissociation for H2X:
                        H2X +H2O ↔ HX + H3O
initial                0.15                     0      0
change             -X                     +X      +X
at equlibrium 0.15-X                  X        X
because Ka1 is small we can assume neglect x in H2X concentration
     Ka1      = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15) 
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
        HX + H2O↔ X^2 + H3O
    8.2x10^-4          Y         8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2       = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m


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