Calculate the temperature at which 0.01% (w/V) aqueous solution of glucose will exhibit the osmotic pressure of 16.4 atm (RMM of glucose=180 g/mol) a. 3.6 K b. 3500K c. 360 K d. 36 K
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To do this, we need to convert the mass percent concentration to molarity. 0.01% (w/V) means that there is 0.01g of glucose in 100mL of solution. To convert this to moles, we divide by the molar mass of glucose: 0.01g / 180g/mol = 0.0000556 mol Show more…
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