To prepare 750ml of 6.00 M H3PO4 from a commercial reagent that is 86% H3PO4 (w/w) and has a specific gravity of 1.71, we first need to calculate the amount of actual H3PO4 present in the reagent.
We know that the commercial reagent is 86% H3PO4 (w/w), which means that in 100g of the reagent, 86g is H3PO4. Since the specific gravity of the reagent is 1.71, this means that 1 ml of the reagent weighs 1.71g.
So, for 100ml of the reagent:
- Mass of H3PO4 = 86g
- Volume of H3PO4 solution = 100ml
- Volume of reagent = 100ml x 1.71 = 171g
Now, we can calculate the molarity of the H3PO4 in the reagent:
(86g H3PO4 / 98g/mol) / 0.171L = 5.64 M
To prepare a 6.00 M H3PO4 solution, we need to dilute the commercial reagent with water. To make 750ml of 6.00 M H3PO4, we calculate the volume of the reagent needed using the formula:
M1V1 = M2V2
(5.64 M) x V1 = (6.00 M) x 0.750 L
V1 = (6.00 M * 0.750 L) / 5.64 M
V1 = 0.797 L
Therefore, we need 0.797L of the commercial reagent to prepare 750ml of 6.00 M H3PO4. This can be diluted with water to reach the desired volume.
Density and Specific Gravity of Solutions: Write the complete solutions:
Describe the preparation of 750ml of 6.00 M H3PO4 from the commercial reagant that is 86% H3PO4 (w/w) and has a specific gravity of 1.71
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