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Nadya [2.5K]
3 years ago
11

A sucrose solution is prepared to a final concentration of 0.250 M . Convert this value into terms of g/L, molality, and mass %.

(Use the following values: molecular weight MWsucrose = 342.296 g/mol ; density rhosol′n = 1.02 g/mL ; and mass of water, mwat = 934.4 g ). Note that the mass of solute is included in the density of the solution.
Chemistry
1 answer:
MArishka [77]3 years ago
6 0

Answer:

A. 85.6 g

= 0.0856 kg.

B. 0.00027 mol/g

= 0.27 mol/kg.

C. 8.39 %

Explanation:

Given:

Molar concentration = 0.25 M

Molar weight of sucrose = 342.296 g/mol

Density of solution = 1.02 g/mL

Mass of water = 934.4 g.

Density in g/l = 1.020 g/ml * 1000ml/1 l

= 1020 g/l

Mass of solution in 1 l of solution = 1020 g

Mass of solution = mass of solvent + mass of solute

Mass of sucrose = 1020 - 934.4

= 85.6 g of sucrose in 1 l of solution.

A.

Density of sucrose = mass/volume

= molar mass/molar concentration

= 342.296 * 0.25

= 85.6 g/l

Number of moles = mass/molar mass

= 85.6/342.296

= 0.25 mol

B.

Molality = number of moles of solute/mass of solvent

= 0.25/934.4

= 0.00027 mol/g

C.

% mass of sucrose = mass of sucrose/total mass of solution * 100

= 85.6/1020 * 100

= 8.39 %

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HCOONa    ⇒     Na⁺ +    HCOO⁻

Consider the ICE take for the formate  ion as:

                                   HCOO⁻ + H₂O   ⇄   HCOOH + OH⁻

At t=0                              0.35                            -              -

At t =equilibrium           (0.35-x)                          x           x            

The expression for dissociation constant of sodium formate is:

K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}

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Chlorine pentafluoride gas is collected at -17.0 °C in an evacuated flask with measured volume of 35.0 L. When all the gas has b
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Answer:

1. The mass of Chlorine pentafluoride, ClF5, is 39.16g

2. The number of mole of Chlorine pentafluoride, ClF5, is 0.3mole

Explanation:

1. To solve the mass of Chlorine pentafluoride, ClF5, first, let us calculate the molar mass of ClF5. This is illustrated below:

Molar Mass of ClF5 = 35.5 + (5 x 19) = 35.5 + 95 = 130.5g/mol

From the ideal gas equation:

PV = nRT (1)

Recall:

Number of mole(n) = mass (m) /Molar Mass(M)

n = m/M

Now substituting the value of n in equation 1, we have:

PV = nRT

PV = mRT/M

Now we can obtain the mass of Chlorine pentafluoride ClF5 as follow:

Data obtained from the question include:

T (temperature) = -17.0 °C = - 17 + 273 = 256K

V (volume) = 35L

P (pressure) = 0.180 atm

R (gas constant) = 0.082atm.L/Kmol

m (mass of Chlorine pentafluoride) =?

M (molar mass of Chlorine pentafluoride) = 130.5g/mol

PV = mRT/M

0.180 x 35 = m x 0.082 x 256/ 130.5

Cross multiply to express in linear form as shown below:

m x 0.082 x 256 = 0.180x35x130.5

Divide both side by 0.082 x 256

m = (0.180x35x130.5) /(0.082x256)

m = 39.16g

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Mass of ClF5 = 39.16g

Molar Mass of ClF5 = 130.5g/mol

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Number of mole = Mass /Molar Mass

Mole of ClF5 = 39.16/130.5g

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