Question #122142
Describe the preparation of the following solution (show all calculations): 5.0 L of 6.000 M H3PO4 from a concentrated solution that is 85.1%(m/m) and has a specific gravity of 1.69
1
Expert's answer
2020-06-15T14:43:01-0400

In order to prepare VV= 5 L of cc = 6.000 M H3PO4 solution, one needs the following quantity nn of H3PO4:

n(H3PO4)=cV=56.000=30n(H_3PO_4) = cV=5·6.000 = 30 mol.

The mass of this quantity of H3PO4 can be calculated using the molar mass MM of H3PO4 (98.00 g/mol):

m=nM=3098.00=2940m = nM = 30·98.00 =2940 g.

Thus, the mass of the concentrated solution needed contains 2940 g of H3PO4:

ms=29400.851=3454.8m_s = \frac{2940}{0.851} = 3454.8 g.

The specific gravity of the concentrated solution is 1.69 g/mL. Therefore, 3454.8 g of the concentrated solution corresponds to the following volume:

Vs=md=3454.81.69=2044V_s = \frac{m}{d} = \frac{3454.8}{1.69}=2044 mL.

Then, to 2044 mL of the concentrated solution one needs to add the following volume of water:

Vw=50002044=2956V_w = 5000-2044 = 2956 mL.

Finally, in order to prepare 5 L of 6.000 M of H3PO4 from a concentrated solution

that is 85.1%(m/m) and has a specific gravity of 1.69 g/mL, one needs to take 2044 mL of the concentrated solution and add 2956 mL of water.


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