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1 & 2 please
1) Next weeks lab requires the TA to make 5.0 L of 0.10 M NaH PO, and 5.0 L of 0.10 M of Na HPO, using the dry forms of the
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Answer #1

Hello, i am sorry to say that as per Chegg guidelines experts can answer either only one full question that too first question or first 4 sub-parts like a), b), c) & d) only. I request you to please post remaining question as separate one since my allotted time could not permit me to answer more than one question also. Thank you in advance for your continued support to us.


Moreover, Table 1 is also not provided if i have enough time to attempt second question, i'm sorry for that.

we know molarity = moles / liter; moles = [mass of substance / molar mass of substance] / liter.

Formula we have to use is

mass of substance in g = (Molarity x molar mass of substance in g/mol) x volume in liters

a)

mass of dry NaH2PO4 in grams = (0.10 M x 119.9770 g/mol ) x 5.0 L = 59.9885 g

so mass of dry NaH2PO4 required in grams = 60 g

b)

mass of dry Na2HPO4 in grams = (0.10 M x 141.9588 g/mol ) x 5.0 L = 70.9794 g

so mass of dry Na2HPO4 required in grams = 71 g

Hope this helped you!

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