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crimeas [40]
3 years ago
9

Which of the following 0.820 M solutions would have the greatest colligative effect?

Chemistry
1 answer:
eimsori [14]3 years ago
8 0

Answer:

K3PO4

Explanation:

Recall that colligative properties depends on the number of particles present. The greater the number of particles present, the greater the degree of colligative properties of the solution. Let us look at each option individually;

SrCr2O7-------> Sr^2+ + Cr2O7^2- ( 2 particles)

C4H11N (not ionic in nature hence it can not dissociate into ions)

K3PO4-------> 3K^+ + PO4^3- (4 particles)

Rb2CO3-------> 2Rb^+ + CO3^2- (3 particles)

Hence K3PO4 has the greatest number of particles and will display the greatest colligative effect.

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112 g of aluminum carbide react with 174 g water to produce methane and aluminum hydroxide in the reaction shown below.
dolphi86 [110]

<u>Answer:</u> 4.999 moles of excess reactant will be left over.

<u>Explanation:</u>

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}       .....(1)

Given mass of aluminium carbide = 112 g

Molar mass of aluminium carbide = 143.96 g/mol

Putting values in equation 1:

\text{Moles of aluminium carbide}=\frac{112g}{143.96g/mol}=0.778mol

For the given chemical reaction:

2Al_4C_3(s)+12H_2O(l)\rightarrow 3CH_4(g)+4Al(OH)_3(s)

By the stoichiometry of the reaction:

2 moles of aluminium carbide reacts with 12 moles of water

So, 0.778 moles of aluminium carbide will react with = \frac{12}{2}\times 0.778=4.668 mol of water

Given mass of water = 174 g

Molar mass of water = 18 g/mol

Putting values in equation 1:

\text{Moles of water}=\frac{174g}{18g/mol}=9.667mol

Moles of excess reactant (water) left = 9.667 - 4.668 = 4.999 moles

Hence, 4.999 moles of excess reactant will be left over.

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