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Calculate The Molarity:
a.___ 3.5 moles of NaNO3 in 4 liters of solution.
b.___4.8 moles of NaOH in 3400mL of solution
c.___30.9 grams of NaOH in 400 mLs of solution (i. molar mass=40g/mol)
d.___340 grams of Na2CO3 in 2 liters of solution (i.molar mass = 105.99 g/mol
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Respuesta :

SvetkaChem

Answer:

a) 0.88 M, b) 0.88 M, c) 1.93 M, d) 1.6 M

Explanation:

a) Molarity = number moles/L solution = 3.5 mol/ 4 L = 0.875mol/L ≈ 0.88 M

b) 3400 mL = 3.4L

Molarity = number moles/L solution = 4.8 mol/ 3.4 L ≈ 1.4 mol/L = 0.88 M

c) 30.9 g * 1 mol/40 g = 30.9/40 mol

400 mL = 0.4 L

Molarity = number moles/L solution = (30.9/40) mol/ 0.4 L ≈ 1.93  mol/L =

= 1.93 M

d) 340 g * 1 mol/105.99 g = 340/105.99 mol

Molarity = number moles/L solution = (340/105.99) mol/ 2L ≈ 1.6 mol/L =

= 1.6 M

The molarity of a is 0.88 M, b is 1.41 M, c is 1.93 M and d is 1.6 M.

Calculation of molarity:

The concentration of a solution calculated as the number of moles of solute per liter of solution is termed molarity.

Molarity can be determined by using the formula,

M = Number of moles/Liters of solution

a. The number of moles of NaNO3 is 3.5 moles, the volume of the solution is 4 L. Now putting the values we get,

[tex]M = \frac{3.5 moles}{4 L} \\M = 0.88 M[/tex]

b. The number of moles of NaOH is 4.8 moles, and the volume of the solution is 3.4 L. Now putting the values we get,

[tex]M = \frac{4.8 moles}{3.4 L} \\M = 1.41 M[/tex]

c. The mass of NaOH is 30.9 grams, the volume of the solution is 400 ml of 0.4 L. The molecular mass of NaOH is 40 g/mol. The number of moles of NaOH will be,

Number of moles = Weight/Molecular mass

Now putting the values we get,

[tex]n = \frac{30.9 g}{40 g/mol} \\n = 0.77 moles[/tex]

Now the molarity will be,

[tex]M = \frac{0.77 moles}{0.4 L} \\M = 1.93 M[/tex]

d. The mass of Na2CO3 given is 340 grams. The volume of the solution is 2 L. The molecular mass of sodium carbonate is 105.99 g/mol. The number of moles of Na2CO3 will be,

[tex]n = \frac{340 g}{105.99 g/mol} \\n = 3.21 moles[/tex]

Now the molarity will be,

[tex]M = \frac{3.21 moles}{2L} \\M = 1.6 M[/tex]

Thus, the molarity of a is 0.88M, b is 1.41 M, c is 1.93 M and d is 1.6 M.

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