A student has a bottle labeled 0.750% m/V glucose solution. The bottle contains

Question

A student has a bottle labeled 0.750% m/V glucose solution. The bottle contains exactly 25.0 mL How much water must the student add to nuke the glucose concentration become 0.100 % m/V? The lab manager has 115 mL of 12.3 mg. ml. and dilutes it to a final volume of 680. mL What is the concentration of the diluted solution? Two solutions. A and B. ard separated by a semipermeable membrane. For each case, predict whether there will be a net flow of water in one direction and. if so, which direction. A with 0.20 M NaCI and solution B with 0.15 M Na_2SO_4 B with 0 10 M K,P04 and solution B with 0 40 M C_6H_12O_6 Which solution has the lower freezing point? 0.25 m KCI or 0.15 m Na_3PO_4 Calculate the number of eq/L PO_4 in a solution that 1.3 Times 10^-3 M inPO_4^2. Sketch the interaction of a water molecule with a potassium ion 11 Sketch the interaction of a water molecule with a bromide ion. What type of solute dissolves readily in water? What type of solute dissolves readily in gasoline? m Blood is essentially an aqueous solution, but it must transport a variety of nonpolar substances (hormones for example) Colloidal proteins, termed albumins, facilitate this transport. Must these albumins be polar or nonpolar? Why?

Explanation / Answer

5)

M1=0.7505 m?L

V1 = 25 ml

M2 = 0.1005 m/L

Now M1V1= M2V2

so 0.750*25 = 0.100*V2

so V2 = 0.750*25/0.100 = 187.5

additional water to beadded = 187.5- 25 = 162.5 ml

6)

V1 = 115 mL

c1 = 12.3 mg/mL

v2 = 680

c2 = ?

c1v1 = c2v2

so 12.3*115 = c2* 680

so c2 = 12.3*115/680 = 2.080 mg/mL

7)

first case

= 1 + 2 =c1Rt +c2RT   = (c1 +c2) RT = (0.20 + 0.15) RT = 0.35RT

second case

= 1 + 2 =c1Rt +c2RT   = (c1 +c2) RT = (0.10 + 0.40) RT = 0.50RT

so OP in second case is greater than Op in first case

So flow will be from higher Op to lower Op side that is from second case to first case side solution

8)

Depression in freezing point = KF* b

where b is molality

so KCl (0.25 m) has higher depression in freezing point than Na3PO4(molality = 0.15)

9)

(eq /L) / (Mol/L) for pO4-3 = ( mass/molweight/3/L )/(mass/Mol weight /L ) = 3

so eq/L = 3 * mol/L PO4-3    = 3*1.3*10^-3 = 3.9 *10^-3

10)

          + H - O - H +

                   -2.......K+

11)  

+ H - O -    H + ......Br-
                   -2..

12) Polar solute dissolves in polar H2) readily

13) Nonpolar solute dissolves in nonpolar gasoline readily

14)

   Blood is non polar

        

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