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crimeas [40]
2 years ago
9

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

Chemistry
1 answer:
eimsori [14]2 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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A helium balloon with a volume of 5.0 L and a pressure of 105 kPa at 25°C floats up high in the sky where the pressure is 850 mm
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Calculate the molality of isoborneol in the product if, a) the melting point of pure camphor is 179°C and the melting point take
tresset_1 [31]

Answer:

The molality of isoborneol in camphor is 0.53 mol/kg.

Explanation:

Melting point of pure camphor= T =179°C

Melting point of sample = T_f = 165°C

Depression in freezing point = \Delta T_f=?

\Delta T_f=T- T_f=179^oC-165^oC=14^oC

Depression in freezing point  is also given by formula:

\Delta T_f=i\times K_f\times m

K_f = The freezing point depression constant

m = molality of the sample

i = van't Hoff factor

We have: K_f = 40°C kg/mol

i = 1 (  organic compounds)

\Delta T_f=14^oC

14^oC=1\times 40^oC kg/mol\times m

m=\frac{14^oC}{1\times 40^oC kg/mol}=0.35 mol/kg

The molality of isoborneol in camphor is 0.53 mol/kg.

8 0
2 years ago
What is the mass of 4.5 x 10^22 molecules of hydrogen peroxide (H2O2)? Show your work in the space below.
valentinak56 [21]

Given :

Number of molecules of hydrogen peroxide, N = 4.5 × 10²².

To Find :

The mass of given molecules of hydrogen peroxide.

Solution :

We know, 1 mole of every compound contains Nₐ = 6.022 × 10²³ molecules.

So, number of moles of hydrogen peroxide is :

n = \dfrac{N}{N_a}\\\\n = \dfrac{4.5\times 10^{22}}{6.022\times 10^{23}}\\\\n = 0.0747 \ moles

Now, mass of hydrogen peroxide is given as :

m = n × M.M

m = 0.0747 × 34 grams

m = 2.54 grams

Hence, this is the required solution.

6 0
2 years ago
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