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Post by Admin Sat Sep 26, 2015 10:48 pm

01) How would you prepare 400 ml of a 0.24 M NaCl solution (MW of NaCl= 58.44 g/mole)?

02) How would you prepare 750 ml of a 0.35 M Na2PO4 solution (MW of Na2PO4 = 141.96 g/mole)?

03) What volume of stock 0.35 M Na2PO4 is needed to prepare 150 ml of 0.2 M Na2PO4?

04) What volume of stock 0.15 M citric acid is needed to prepare 230 ml of 2.5x10‐3 M citric acid?

05) If you have 35 ml of 0.1 M NaCl, is it enough to prepare 200 ml of 0.06 M NaCl?  
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Post by Ravindee Mon Oct 26, 2015 7:31 pm

1)[V x C x MW) 0.4 L x 0.24 moles/L x 58.44 g/mole = 5.61 g NaCl dissolved in 400 ml water.

2)0.75 L x 0.35 moles/L x 141.96 g/mole = 37.265g of Na2PO4 dissolved in 750 ml water.

3)C1V1 =C2V2
V1=(0.2 M)(150 ml)/ (0.35 M) = 85.714 ml of stock Na2PO4.

4)V1=(0.0025 M)(230 ml)/ (0.15 M) = 3.833 ml of stock citric acid.

5)[cv=cv] solve for V1: V1 = ((0.06 M)(200 ml)/(0.1 M) = 120 ml, so no, you do not have enough stock.
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