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The solubility of CO gas in water is 0.158/(100 mL) at a CO, pressure of 760 mmHg Review You may want to reference Pages 259-
Part A Mass of glucose needed to prepare 305.0 mL of 24 % w/v glucose (C H1206). Express your answer using two significant fi
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Answer #1

1) Solubility of CO2 in water = 0.15 g/100 ml

Moles of CO2 = Mass/molar mass = 0.15/44 = 0.00341 moles

Volume = 100 ml = 0.1 liter

Solubility of CO2 in Mol/Lit = S = 0.00341/0.1 = 0.0341 M

We know that;

S = K * P

S is molar solubility , K is Henry constant for CO2 and P is pressure.

P = 760 mmHg = 1 atm

0.0341 = K * 1

K = 0.0341 M/atm

Again apllying same equation;

S' = K * P

S' = 0.0341 * 4.1 = 0.1398 M = 0.14 Mol/Liter ....Answer

2)

24% (w/v) means 100 ml solution contains 24 g of glucose.

100 ml solution contains 24 g glucose

So, 305 ml solution will contain = 24 * 305/100 = 73.2 grams

So, mass of glucose = 73.2 g OR 73 g (approx) ....Answer

Let me know if any doubt.

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