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liberstina [14]
2 years ago
11

What is the number of moles of solute in a 0.3 molal solution containing 0.10 kg of solvent? Don't forget the unit.

Chemistry
2 answers:
timurjin [86]2 years ago
5 0
Molality is one way of expressing concentration for solutions. It has units of moles of solute per kg of solvent. From the given values, we easily calculate for the moles of solute by multiplying the mass of solvent to the molality. We do as follows:

moles solute = 0.3 (10) = 3 mol solute


jeka57 [31]2 years ago
5 0

Answer:

The number of moles of solute in a 0.3 molal solution containing 0.10 kg of solvent is 0.03.

Explanation:

Molality (m) is the number of moles of solute that are dissolved in 1 kilogram of solvent.

Molality is then determined by the expression:

Molality (m)=\frac{number of moles of solute}{kilograms of solvent}

Molality is expressed in units \frac{moles}{kg}.

Then you can apply a rule of three as follows to calculate the amount of moles of solute: if, according to the given molality, 1 kg of solution has 0.3 moles of solute, 0.10 kg of solution how many moles of solute are there?

moles=\frac{0.10 kg*0.3 moles}{1 kg}

moles=0.03

The number of moles of solute in a 0.3 molal solution containing 0.10 kg of solvent is 0.03.

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0.9775 grams of an unknown compound is dissolved in 50.0 ml of water. Initially the water temperature is 22.3 degrees Celsius. A
elena-14-01-66 [18.8K]

Answer:

The enthlapy of solution is -55.23 kJ/mol.

Explanation:

Mass of water = m

Density of water = 1 g/mL

Volume of water = 50.0 mL

m = Density of water × Volume of water = 1 g/mL × 50.0 mL=50.0 g

Change in temperature of the water ,ΔT= 27.0°C - 22.3°C = 4.7°C

Heat capacity of water,c =4.186 J/g°C

Heat gained by the water when an unknown compound is dissolved be Q

Q= mcΔT

Q=50.0 g\times 4.186 J/g^oC\times 4.7^oC=983.71 J

heat released when 0.9775 grams of an unknown compound is dissolved in water will be same as that heat gained by water.

Q'=-Q

Q'= -983.71 J =-0.98371 kJ

Moles of unknown compound = \frac{0.9975 g}{56 g/mol}=0.01781 mol

The enthlapy of solution :

\frac{Q'}{moles}

=\frac{-0.98371 kJ}{0.01781 mol}=-55.23 kJ/mol

The enthlapy of solution is -55.23 kJ/mol.

8 0
2 years ago
A flexible plastic container contains 0.860g of helium has in a volume of 19.2L if 0.205g of helium is removed at contact pressu
mash [69]

<u>Given:</u>

Initial volume of He, V1 = 19.2 L

Initial mass of He, m1 = 0.0860 g

Mass of He removed = 0.205 g

<u>To determine:</u>

The new volume of He i.e V2

<u>Explanation:</u>

Based on Avogadro's law:

Volume of a gas is directly proportional to the # moles of the gas

Volume (V) α moles (n) -----(1)

Atomic mass of He = 4 g/mol

Initial moles of He, n1 = 0.860 g/4 g.mol-1 = 0.215 moles

Final moles of He, n2 = (0.860-0.205)g/4 g.mol-1 = 0.164 moles

Based on eq(1) we have:

V1/V2 = n1/n2

V2 = V1 n2/n1 = 19.2 L * 0.164 moles/0,215 moles = 14.6 L

Ans: New volume is 14.6 L



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