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Directions: Answer the following sets of problems in the space provided. Show your solution.

1. In a solution, there is 120.0 mL (125.605 g) solvent and 6.24 mL (7.0508 g) solute present in a solution. Find the mass percent and volume percent.

2. A 1500 mL solution is composed of 0.25 mol NaCl dissolved in water. Find its molarity.

3. What is the concentration in part per million when a 10 g CaCl is dissolved in 100 mL solution.

4. How many moles of CaCO3 is present in a 0.05 molality solution with 750 kg solvent.

5. Determine the mole fraction of 25 g NaCl dissolved in 100 grams H20. (Molar mass: NaCl = 58 g/mol; H20 = 18 g/mol)​

Sagot :

RICHARDABELAIN

Answer:

2)As Molarity= moles of solute/ volume of solution(liters)

M= 5.85*1000/((23+35.5)*500)=0.2M

3) (molarity) = moles of solute/ L of solution

M= 0.2567mols/0.25L = 1.03 mol/L or 1.03M

4)moles of NaOH = 10 g NaOH ×

1

mol NaOH

40.00

g NaOH

= 0.25 mol NaOH

kilograms of H₂O = 500 g H₂O ×

1

kg H₂O

1000

g H₂O

= 0.500 kg H₂O

molality =

moles of solute

kilograms of solvent

=

0.25

mol

0.500

kg

= 0.50 mol/kg

5)

1

kg water

⋅

0.428 moles NaCl

20

kg water

=

0.02 moles NaCl

Explanation:

Yan lang po alam ko, kayo na po bhala sa no 1)

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