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Unformatted text preview: is fasting has blood levels of 0.00444 M glucose (C6H12O6, 180.2 g/mol). What is the percent glucose, by mass, in an aqueous 0.00444 M glucose solution whose density is 1.003 g/mL? 0.00444 mol C6H12O6 180.2 g C6H12O6 0.00444 M C6H12O6 = 0.00444 mol C6H12O6 1 L solution 1 mol C6H12O6 = 0.800088 g C6H12O6 +2 g glucose 1 L solution 1000 mL soln 1.003 g soln = 1003 g soln 1 L soln 1 mL soln +2 g solution 0.800088 g C6H12O6 100% = 0.0798% glucose 1003 g soln +1 answer +1 SF +2 mass % 8 (b) If you had a 0.00444 M glucose (C6H12O6) solution and a 0.00250 M saline (NaCl) solution, which solution would cause a higher osmotic pressure at the same T? Describe, at the molecular level, why. i = 2 for NaCl while i = 1 for glucose, so iM for glucose = 0.00444 M while iM for NaCl = 0.00500 M
+2 correct i values Since = iMRT, the higher osmotic pressure correlates to the higher iM value. The saline solution would cause a higher osmotic pressure. +1 NaCl solution based on reasoning Colligative properties are more influenced by larger number o...
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This note was uploaded on 09/20/2012 for the course CHEM 1252 taught by Professor Jew during the Fall '12 term at UNC Charlotte.
- Fall '12