Ement 15: Colligative Properties Experiment 15: Colligative Properties 7. Solution preparation: a. How many grams of NaCl will you need to prepare 50.0 mL of 1.00M NaCI? b. How many grams of NaCl will you need to prepare 50.0 mL of 3.00M NaCl? c. How many grams of glycerol will you need to prepare 50.0 mL of 1.00M glycerol (C3H3O3)? d. How many grams of glycerol will you need to prepare 50.0 mL of 3.00M glycerol (C3H$O3)?

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Ement 15: Colligative Properties
Experiment 15: Colligative Properties
7. Solution preparation:
a. How many grams of NaCl will you need to prepare 50.0 mL of 1.00M NaCI?
b. How many grams of NaCl will you need to prepare 50.0 mL of 3.00M NaCl?
c. How many grams of glycerol will you need to prepare 50.0 mL of 1.00M glycerol
(C3H3O3)?
d. How many grams of glycerol will you need to prepare 50.0 mL of 3.00M glycerol
(C3H$O3)?
Transcribed Image Text:Ement 15: Colligative Properties Experiment 15: Colligative Properties 7. Solution preparation: a. How many grams of NaCl will you need to prepare 50.0 mL of 1.00M NaCI? b. How many grams of NaCl will you need to prepare 50.0 mL of 3.00M NaCl? c. How many grams of glycerol will you need to prepare 50.0 mL of 1.00M glycerol (C3H3O3)? d. How many grams of glycerol will you need to prepare 50.0 mL of 3.00M glycerol (C3H$O3)?
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Since we only answer up to 3 sub-parts, we’ll answer the first 3. Please resubmit the question and specify the other subparts (up to 3) you’d like answered.

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Stoichiometry is mainly based on the calculation of moles and volumes. These two values are used to calculate the molarity of solution. The relation between moles, and mass with molarity is as given below;

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Step 3

Part (a) 

Volume – 50 mL  = 50 x 10-3 L

Molarity = 1.00 M

Molar mass of NaCl = 58.5 g/mol

Substitute the values to calculate mass:

Chemistry homework question answer, step 3, image 1

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