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Which of the following solutions contains the largest number of moles of dissolved particles? a.) 25 ml of sodium chloride b.) 25 ml of aluminum hydroxide c.) 25 ml of acetic acid d.) 25 ml of nitric acid e.) 25 ml of sugar

User Monalisa
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2 Answers

6 votes

Answer:

b.) 25 ml of aluminum hydroxide.

Step-by-step explanation:

Hello,

In this case, bu knowing the formula of each substance, we can predict the number of particles are going to be dissolved in the given solutions, thus we have:

Sodium chloride, NaCl: when it is placed in water, two ions, sodium and chloride are dissolved.

Aluminium hydroxide, Al(OH)3: when it is placed in water, four ions are dissolved, one aluminium (III) and three hydroxyl.

Acetic acid CH3COOH: when it is placed in water, and equilibrium dissolution is present, therefore, not all the acetic acid particles will be dissolved as ions, anyway, two ionic species are formed, acetate anion and hydronium.

Sugar C12H22O11: when it is placed in water, it is not properly dissociated but it is said to form hydrogen bonds in order to get dissolved in water, therefore, just one dissolved particle is formed.

In such a way, the substance having the larges number of moles of dissolved particles is aluminium hydroxide having four, by considering the same concentration for all the solutions, say 1 M or that so.

Regards.

User David Cheung
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5 votes

Answer:

b.) 25 ml of aluminum hydroxide

Step-by-step explanation:

For this question, we have to assume that we have the same concentration for all the solutions, for example, 1 M. Additionally, we have to take into account the ionization reaction for each species:

a)
NaCl~->~Na^+~+~Cl^- we have to ions

b)
Al(OH)_3~->~Al^+^3~+~3OH^- we have fourth ions

c)
CH_3COOH~->~CH_3COO^-~+~H^+ we have two ions

d)
C_6H_1_2O_6~->~C_6H_1_2O_6 we have one ion

If we have the same volume and the same concentration the variables that will help us to answer the question would be the number of ions. If we have more ions we will have more particles dissolved. Therefore the answer would be b) (due to the fourth ions).

I hope it helps

User Thuga
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