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Question

In Day 1 of the Kinetics Lab, you will start by mixing a stock solution of Methyl Orange that you will then dilute 5 times in order to have a set of solutions to test the response of your colorimeters. In this problem you will calculate the [MO] in your stock solution that you will draw from to make your 5 diluted solutions. To make the stock solution, you add 3.10 mL of deionized water, 3.04 mL of 2.0 M HCl, and 4.07 mL of 139 g/L Methyl Orange into your graduated cylinder. Assuming that the molar mass of MO is 327.33 g/mol, what is the [MO] in this mixture in units of molarity?

Answer 1

Given:

139g/L methyl orange 

1L contains 139g methyl orange

1000 ml contains 139 g methyl orange 

1ml contains 139/1000 g methyl orange

4.07 ml contains 139/1000 x 4.07 g methyl orange

4.07 ml contains 0.566 g methyl orange 

3.10 ml deionized water + 3.04 ml of 2M HCl + 4.07 ml of methyl orange 

Total volume is 10.21 ml = 0.01021 litre

Molarity = number of moles of methyl orange/volume of solution in liters

number of moles = mass of methyl orange / molecular weight of MO = 0.566 / 327.33 = 0.00173 moles

Molarity = 0.00173 / 0.01021 = 0.169 M = 169 mM

Hence the Molarity [MO] in this mixture is 169 mM

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