Dealing with a 0.150 M Co (NO3) 2•6H2O. Determine the number of grams of Co (NO3) 2•6H2O that you would need to weigh out in order to prepare 100.00 mL of the solution

Question

Chemistry

Maximillian Steele

10 March, 19:11

Dealing with a 0.150 M Co (NO3) 2•6H2O. Determine the number of grams of Co (NO3) 2•6H2O that you would need to weigh out in order to prepare 100.00 mL of the solution

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Answers (2)

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Anastasia Ortega · Answer 1

4.365g

Explanation:

We'll begin by calculating the number of mole of Co (NO3) 2•6H2O present in 100mL of 0.150 M Co (NO3) 2•6H2O solution. This is illustrated below:

Molarity = 0.150M

Volume of solution = 100mL

Recall: 1000mL = 1L

Therefore, 100mL = 0.1L

Mole of solute (Co (NO3) 2•6H2O) = ?

Molarity is simply mole per unit litre of solution. It is represented mathematically as:

Molarity = mole / Volume

Mole = Molarity x Volume

Mole of Co (NO3) 2•6H2O = 0.15x0.1

Mole of Co (NO3) 2•6H2O = 0.015mol

Now, we need to convert 0.015mol to grams in order to obtain the desired result. This is illustrated below:

Mole of Co (NO3) 2•6H2O = 0.015mol

Molar Mass of Co (NO3) 2•6H2O = 59 + 2[14 + (16x3) ] + 6[ (2x1) + 16] = 59 + 2[14 + 48] + 6[2 + 16] = 59 + 2[62] + 6 [18] = 59 + 124 + 108 = 291g/mol

Mass of Co (NO3) 2•6H2O = ?

Mass = number of mole x molar Mass

Mass of Co (NO3) 2•6H2O = 0.015 x 291

Mass of Co (NO3) 2•6H2O = 4.365g

Therefore, 4.365g of Co (NO3) 2•6H2O is needed to prepare 100mL of 0.150 M Co (NO3) 2•6H2O solution.

Jalen Keller · Answer 2

The correct answer is 4.3635 g

Explanation:

First, we have to calculate the molecular weight of the solute (Co (NO₃) ₂•6H₂O). For this, we obtain the molar mass of the elements from the Periodic Table (Co, N, O and H) and calculate as follows:

MW (Co (NO₃) ₂•6H₂O) = (1 x MM (Co)) + (2 x MM (N)) + ((2 x 3) x MM (O)) + ((6 x 2) x (MM (H)) + (6 x (MM (O)))

= (58.9 g/mol) + (2 x (14 g/mol)) + (6 x 16 g/mol) + (12 x 1 g/mol) + 6 x (16 g/mol)

= 290.9 g/mol

We have a solution 0.150 M Co (NO₃) ₂•6H₂O, so we can convert from mol/L to g/L if we multiply the concentration by the molecular weight as follows:

0.150 mol/L x 290.9 g / 1 mol = 43.635 g/L

Finally, we have to prepare a 100 ml solution. This volume is equal to 0.1 L, so we multiply the concentration in g/L by the volume in L and we obtain the mass we need to weigh:

mass = 43.635 g/L x 0.1 L = 4.3635 g