A solution of Cuso, is labelled 1.743 M. How much Cuso, in grams, must be used to make 1.4957 liter of solution?
Select one:
O a. 416.1
Ob. 0.007301
O c. 166.6
Od. 129.2

Option A. 416.1 g.

Explanation:

Data obtained from the question include the following:

Molarity = 1.743 M

Volume = 1.4957 L

Mass of CuSO4 =..?

Next, we shall determine mole of CuSO4 present in the solution.

This is illustrated below:

Molarity is simply defined as the mole of solute per unit litre of the solution. Mathematically, it is expressed as:

Molarity = mole /Volume

With the above formula, we can obtain the mole CuSO4 present in the solution as follow:

Molarity = 1.743 M

Volume = 1.4957 L

Mole of CuSO4 =..?

Molarity = mole /Volume

1.743 = mole of CuSO4 /1.4957

Cross multiply

Mole of CuSO4 = 1.743 x 1.4957

Mole of CuSO4 = 2.607 moles

Finally, we shall convert 2.607 moles of CuSO4 to grams.

This can be obtained as follow:

Molar mass of CuSO4 = 63.55 + 32.06 + (16x4) = 159.61 g/m

Mole of CuSO4 = 2.607 moles

Mass of CuSO4 =..?

Mole = mass /molar mass

2.607 = mass of CuSO4 /159.61

Cross multiply

Mass of CuSO4 = 2.607 x 159.61

Mass of CuSO4 = 416.1 g

Therefore, 416.1 g of CuSO4 is needed to prepare the solution.

the millers tale is the answer.

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"