The amount (in grams) of potassium dichromate (MW = 294) present in 75 mL of 0.16 M aqueous solution is _______Correct answer is between '3.52,3.54'. Can you explain this answer?

Question

Answer

Calculation of Amount of Potassium Dichromate in Solution

Given:
- Volume of solution = 75 mL
- Concentration of solution = 0.16 M
- Molecular weight of potassium dichromate = 294 g/mol

Formula:
- Amount of solute (in grams) = Concentration × Volume × Molecular weight

Solution:

- Convert volume from milliliters to liters:
- 75 mL = 0.075 L

- Substitute given values into the formula:
- Amount of solute (in grams) = 0.16 M × 0.075 L × 294 g/mol
- Amount of solute (in grams) = 3.528 g

- Round off the answer to two decimal places:
- Amount of potassium dichromate present in 75 mL of 0.16 M aqueous solution is 3.54 g.

Therefore, the correct answer is between 3.52 and 3.54 g.

Answer

No.of .moles in 0.075L of k2cr2o7 = molarity ×volume
= 0.16 × 0.075
= 0.012

=no.of.moles = wt in g/molecular weight
=0.16 =wt in g /294
wt in g = 0.16×294= 3.528

answer(3.528)

Answer

Molarity= no. of moles of solute/vol. of soln.
=given mass of solute/molar mass/vol. of soln.
=0.16*0.075*294
=3.528

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