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monitta
2 years ago
11

What mass of solid that has a molar mass 89.0 g/mol should be added to 100.0 g of benzene to raise the boiling point of benzene

by 2.42°C? (The boiling point elevation constant of benzene is 2.53°C•kg/mol.) 2.84 g 8.52 g 21.5 g 93.0 g
Chemistry
1 answer:
taurus [48]2 years ago
3 0

On the basis of given data:

Mass of solvent (benzene) = 100 gm or 0.1 Kg

Molar mass of solute (solid) = 89 g/mol

Increase in boiling point (deltaTb) = 2.42 degree C

The boiling point elevation constant k = 2.53 C.kg/mol

There is a need to determine the mass of the solid to be added, the elevation in boiling point is proportional to the m (molality) of the solute;

ΔTb = k.m

Here, m that is molality = moles of solute/kg of solvent

Therefore,

ΔTb = k (benzene) × moles of solid/kg of benzene

2.42 = 2.53 × moles of solid/0.1

moles of solid = 0.0956 moles

The molar mass of solid = 89 g/mol

Thus, the mass of solid = 0.0956 moles × 89 g/mol

= 8.508 g

Thus, the mass of solid to be added is 8.508 g.

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so it depends upon

a) how much solvent is present : More the solvent on one side of semipermeable membrane more the movement of solvent particles on the other side of membrane


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2 years ago
Based on the information in the table, which of the following arranges the bonds in order of decreasing
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Calculate a reasonable amount (mass in g) of your unknown acid to use for a titration. You will want about 30 mL of titrant to g
Vinil7 [7]

Answer:

"0.60 g" is the appropriate solution.

Explanation:

The given values are:

Volume of base,

= 30 ml

Molarity of base,

= 0.05 m

Molar mass of acid,

= 400 g/mol

As we know,

⇒  Molarity=\frac{Number \ of \ moles \ of \ base}{Number \ of \ solution}

On substituting the values, we get

⇒           0.05=\frac{Number \ of \ moles \ of \ base}{30\times 10^{-3}}

⇒  Number \ of \ moles \ of \ base=0.05\times 30\times 10^{-3}

⇒                                             =1.5\times 10^{-3}  

hence,

⇒  Moles \ of \ acid=\frac{Mass \ of \ acid}{Molar \ mass \ of \ acid}

On substituting the values, we get

⇒  1.5\times 10^{-3}=\frac{Mass \ of \ acid}{400}

⇒  Mass \ of \ acid=1.5\times 10^{-3}\times 400

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8 0
3 years ago
A 25.00 mL sample of the ammonia solution
musickatia [10]

Answer:

1.634 molL-1

Explanation:

The mol ration between NH3 and HCl is 1 : 1

Using Ca Va / Cb Vb = Na / Nb   where a = acid and b = base

Na = 1

Nb = 1

Ca = 0.208 molL-1

Cb = ?

Va = 19.64 mL

Vb = 25.00mL

Solving for Cb

Cb = Ca Va / Vb

Cb = 0.208 * 19.64 / 25.0

Cb = 0.1634 molL-1 (Concentration of diluted ammonia solution)

Using the dilution equation;

C1V1 = C2V2

Initial Concentration, C1 = ?

Initial Volume, V1 = 25.00 mL

Final Volume, V2 =  250 mL

Final Concentration, C2 = 0.1634 molL-1

Solving for C1;

C1 = C2 * V2 / V1

C1 = 0.1634 * 250 / 25.00

C1 = 1.634 molL-1

3 0
3 years ago
A closed container contains 0.40 moles of argon gas at 25 °C and a pressure of 740 torr. The container is heated to 125 °C and t
Ksivusya [100]

The number of moles of argon that must be released in order to drop.

Solution:

Initial Temperature = 25°c = 298 K

Final Temperature =125 °c = 398 K

Initial Moles (n1) = 0.40 mole

Now,  Using the ideal gas law,

n1T1 = n2T2

0.400×298 = n2 × 398

n2 = 0.299 mol

Moles of Argon released

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= 0.100 mol.

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