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3 years ago

The addition of which substance will decrease the solubility of AgNO3 in water?

Chemistry

2 answers:

eduard3 years ago

8 0

Answer: The correct answer is Option 3.

Explanation:

Common ion effect is defined as the effect which occurs on equilibrium when a common ion (an ion which is already present in the solution) is added to a solution. This effect decreases the solubility of a solute.

In silver nitrate, the ions present in the solution are: $Ag^+\text{ and }NO_3^-$

For the given options:

Option 1: 0.1 M potassium chloride

The ions formed by the ionization of this solution are $K^+\text{ and }Cl^-$ . As, no common ions are present. So, addition of this substance will not decrease the solubility of silver nitrate solution.

Option 2: 0.1 M barium chloride

The ions formed by the ionization of this solution are $Ba^{2+}\text{ and }Cl^-$ . As, no common ions are present. So, addition of this substance will not decrease the solubility of silver nitrate solution.

Option 3: 0.1 M sodium nitrate

The ions formed by the ionization of this solution are $Na^+\text{ and }NO_3^-$ . As, nitrate ions are present which is a common ion. So, addition of this substance will decrease the solubility of silver nitrate solution.

Option 4: 0.1 M sodium carbonate

The ions formed by the ionization of this solution are $Na^+\text{ and }CO_3^{2-}$ . As, no common ions are present. So, addition of this substance will not decrease the solubility of silver nitrate solution.

Hence, the correct answer is Option 3.

sertanlavr [38]3 years ago

5 0

3 is correct because the .1 moles of the NO3 will be better option to decrease the solubility of the Silver Nitrate.

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$M=32 \ \frac{g}{mol} \\
m=10 \ g \\
n=\frac{10 \ g}{32 \ \frac{g}{mol}}=0.3125 \ mol$

1 mole of glucose reacts with 6 moles of oxygen
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$x=\frac{1 \ mol \cdot 0.3125 \ mol}{6 \ mol} = \frac{3125}{60000} \ mol= \frac{5}{96} \ mol$

Now calculate the mass of 5/96 moles of glucose.
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