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wlad13 [49]
3 years ago
8

Which of the following solutes has the greatest effect on the colligative properties for a given mass of pure water? Explain.

Chemistry
1 answer:
Strike441 [17]3 years ago
8 0

Answer:

Option a.

0.01 mol of CaCl₂ will have the greatest effect on the colligative properties, because it has the biggest i

Explanation:

To determine which of the solute is going to have a greatest effect on colligative properties we have to consider the Van't Hoff factor (i)

These are the colligative properties:

ΔP = P° . Xm . i  →  Lowering vapor pressure

ΔT = Kb . m . i  → Boiling point elevation

ΔT = Kf . m . i  →  Freezing point depression

π = M . R . T  →  Osmotic pressure

Van't Hoff factor are the numbers of ions dissolved in the solution. For nonelectrolytes, the i values 1.

CaCl₂ and KNO₃ are two ionic solutes. They dissociate as this:

CaCl₂  → Ca²⁺ + 2Cl⁻

We have 1 mol of Ca²⁺ and 2 chlorides, so 3 moles of ions → i = 3

KNO₃ → K⁺ + NO₃⁻

We have 1 mol of K⁺ and 1 mol of nitrate, so 2 moles of ions → i = 2

Option a, is the best.

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balu736 [363]
Ammonia is formed by a reaction between hydrogen and nitrogen as shown by the equation below.
N2(g) + 3H2(g) = 2NH3(g)
1 mole of ammonia contains 17 g
Therefore 10.78 g of ammonia are equivalent to 10.78/17 = 0.6341 moles
The mole ratio of hydrogen to ammonia is 3 : 2
Therefore, moles of hydrogen used will be  0.6341 × 3/2 = 0.9512 moles
1 mole of hydrogen is equivalent to 2 g
Thus, the mas of hydrogen will be 0.9512 moles × 2 = 1.9023 g
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8 0
2 years ago
If two balls have the same volume, but ball A has twice as much mass as ball B, which one will have the greater density? ball A,
rewona [7]

Density (D) is defined as the amount or mass (m) of a substance present in a unit volume(V). It can be expressed mathematically as:

Density = Mass/Volume

i.e. D = m/V -------(1)

Units: g/cm3

a)

If m1, V1 and D1 = mass, volume  and density respectively of ball A

m2, V2 and D2 = mass, volume and density respectively of ball B

It is given that: V1 = V2; but m1 = 2m2

Based on equation (1) we have:

D1/D2 = (m1/V1)* (V2/m2) = (2m2/V2)*(V2/m2) = 2

Thus, density of ball A is twice that of B.

Ans: Ball A will have a greater density than B

b)

If m1, V1 and D1 = mass, volume  and density respectively of ball C

m2, V2 and D2 = mass, volume and density respectively of ball D

It is given that:

V1 = 3V2 and m2 = 1/3(m1) i.e m1 = 3m2

Therefore,

D1/D2 = (m1/V1)* (V2/m2) = (3m2/3V2)*(V2/m2) = 1

Thus, D1 = D2

Ans: Their densities are equal

c)

If m1, V1 and D1 = mass, volume  and density respectively of ball P

m2, V2 and D2 = mass, volume and density respectively of ball Q

It is given that:

m1 = m2 but V1 = 2V2

Therefore,

D1/D2 = (m1/V1)* (V2/m2) = (m2/2V2)*(V2/m2) = 1/2

Thus, D1 = 1/2(D2)

Ans: Ball Q will have a greater density.

d)

If m1, V1 and D1 = mass, volume  and density respectively of ball X

m2, V2 and D2 = mass, volume and density respectively of ball Y

It is given that:

V1 = 2V2 and m1 = 1/2(m2)

Therefore,

D1/D2 = (m1/V1)* (V2/m2) = ((1/2(m2)/2V2)*(V2/m2) = 1/4

Thus, D1 = 1/4(D2)

Ans: Ball Y will have a greater density.



5 0
3 years ago
Read 2 more answers
Draw the Lewis dot structure for RbIO2. Include all hydrogen atoms and nonbonding electrons. Show the formal charges of all atom
Rzqust [24]

Answer : The Lewis-dot structure of RbIO_2  is shown below.

Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

The given molecule is, RbIO_2

As we know that rubidium has '1' valence electrons, iodine has '7' valence electrons and oxygen has '6' valence electrons.

Therefore, the total number of valence electrons in RbIO_2 = 1 + 7 + 2(6) = 20

As we know that RbIO_2  is an ionic compound because it is formed by the transfer of electron takes place from metal to non-metal element.

7 0
3 years ago
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